NCERT Solutions Class 12 Physics Chapter 4 Moving Charges and Magnetism

Q.1 A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

Ans.

\begin{array}{l}\text{Distance of the point from the wire},\text{r}=\text{2}0\text{cm}=0.\text{2 m}\\ \text{Magnitude of magnetic field at this point is given by}\\ \text{the}\text{relation}\\ \text{B=}\frac{{\mu }_0}{4\pi }\frac{2I}{r}\\ \text{Here},{\mu }_0=\text{Permeability of free space}\\ =\text{4}\pi \times \text{1}{0}^{–\text{7}}{\text{Tm A}}^{–\text{1}}\\ B=\frac{4\pi \times \text{1}{0}^{–\text{7}}\times 2\times 35}{4\pi \times 0.2}=3.5\times \text{1}{0}^{–5}T\\ \therefore \text{The magnitude of the magnetic field at a point}\\ \text{2}0\text{cm from the wire is 3}.\text{5}\times \text{1}{0}^{–\text{5}}\text{T}.\end{array}

Q.2 A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?

Ans.

\begin{array}{l}\text{Distance of the point from the wire},\text{r}=\text{2}0\text{cm}=0.\text{2 m}\\ \text{Magnitude of magnetic field at this point is given by}\\ \text{the}\text{relation}\\ \text{B=}\frac{{\mu }_0}{4\pi }\frac{2I}{r}\\ \text{Here},{\mu }_0=\text{Permeability of free space}\\ =\text{4}\pi \times \text{1}{0}^{–\text{7}}{\text{Tm A}}^{–\text{1}}\\ B=\frac{4\pi \times \text{1}{0}^{–\text{7}}\times 2\times 35}{4\pi \times 0.2}=3.5\times \text{1}{0}^{–5}T\\ \therefore \text{The magnitude of the magnetic field at a point}\\ \text{2}0\text{cm from the wire is 3}.\text{5}\times \text{1}{0}^{–\text{5}}\text{T}.\end{array}

Q.3

$\begin{array}{l}\text{A long straight wire in the horizontal plane carries a current of 50 A in north to south}\\ \text{direction. Give the magnitude and direction of}\overrightarrow{\mathrm{B}}\text{at a point 2.5 m east of the wire.}\end{array}$

Ans.

$\begin{array}{l}\text{Here, \hspace{0.17em}current flowing\hspace{0.17em}through the wire, I = 50 A}\\ \text{Distance of the point from the wire, r = 2.5 m\hspace{0.17em} east\hspace{0.17em}of\hspace{0.17em}wire}\\ \text{Magnitude of magnetic field at that point is given as:}\\ \text{B =}\frac{{\text{μ}}_{\text{0}}\text{2I}}{\text{4}\mathrm{\pi }\text{r}}\\ {\text{Here,\hspace{0.17em}μ}}_{\text{0}}\text{= 4}\mathrm{\pi }{\text{×10}}^{\text{–7}}{\text{TmA}}^{\text{–1}}\\ \text{B =}\frac{\text{4}\mathrm{\pi }{\text{×10}}^{\text{–7}}{\text{TmA}}^{\text{-1}}\text{×2×50 A}}{\text{4}\mathrm{\pi }\text{×2.5 m}}{\text{= 4×10}}^{\text{–6}}\text{\hspace{0.17em}T}\\ \text{The point is located normal to the plane\hspace{0.17em}of\hspace{0.17em}wire at a distance of 2.5 m.}\\ \text{The current flowing\hspace{0.17em}through the wire is\hspace{0.17em}directed vertically downwards.}\\ \therefore \text{According to the Maxwell’s\hspace{0.17em}right hand thumb rule, the direction of the magnetic field}\\ \text{at the given point is vertically upward.}\end{array}$

Q.4 A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?

Ans.

\begin{array}{l}\text{Current carried}\text{by the power line},\text{I}=\text{9}0\text{A}\\ \text{Point lies below the power line at distance},\text{r}=\text{1}.\text{5 m}\\ \therefore \text{Magnetic field at that point is given by the relation},\\ B=\frac{{\mu }_{0}2I}{4\pi r}\\ \text{Here},{\mu }_0=\text{Permeability of free space}\\ =\text{4}\pi \times \text{1}{0}^{–\text{7}}{\text{TmA}}^{–\text{1}}\\ B=\frac{\text{4}\pi \times \text{1}{0}^{–\text{7}}\times 2\times 90}{\text{4}\pi \times 1.5}=1.2\times \text{1}{0}^{–5}T\\ \text{Since}\text{the current is flowing from East to West}\text{and}t\text{he}\\ \text{point is below the power line,}\\ \therefore \text{According to Maxwell}’\text{s right hand thumb rule},\\ \text{the}\text{direction}\text{of}\text{magnetic}\text{field}\text{is}\text{towards}\text{the}\text{south}\text{.}\end{array}

Q.5 A uniform magnetic field of 1.5 T exists in a cylindrical region of radius10.0 cm, its direction parallel to the axis along east to west.
A wire carrying current of 7.0 A in the north to south direction passes through this region.
What is the magnitude and direction of the force on the wire if,
(a) the wire intersects the axis,
(b) the wire is turned from N-S to northeast-northwest direction,
(c) the wire in the N-S direction is lowered from the axis by a distance of 6.0 cm?

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}strength\hspace{0.17em}}\mathrm{of}\mathrm{m}\text{agnetic field},\text{B}=\text{1}.\text{5 T}\\ \text{Radius of cylindrical region},\text{r}=\text{1}0\text{cm}=0.\text{1 m}\\ \text{Current flowing\hspace{0.17em}through the wire passing through the}\\ \text{cylindrical region},\text{I}=\text{7 A}\\ \mathrm{}\left(\text{a}\right)\text{If the wire intersects the axis},\text{then the length of}\\ \text{the wire is the diameter of the cylindrical region}.\\ \text{Thus},\text{l}=\text{2r}=0.\text{2 m}\\ \text{Angle between current and magnetic field},\mathrm{\theta }=\text{9}0\mathrm{°}\\ \text{Magnetic force acting on the wire is given as:}\\ \text{F}=\text{BIl sin}\mathrm{\theta }=\text{1}.\text{5}\times \text{7}\times 0.\text{2}\times \text{sin 9}0\mathrm{°}=\text{2}.\text{1 N}\\ \therefore \text{A force of 2}.\text{1 N acts on the wire vertically}\\ \text{downwards}.\\ \left(\text{b}\right)\text{New length of the wire after turning it to the}\\ \text{North\hspace{0.17em}east}-\text{North\hspace{0.17em}west direction\hspace{0.17em}is\hspace{0.17em}given as}:\\ {\text{l}}_{\text{1}}=\frac{\mathrm{l}}{\text{sin}\mathrm{\theta }}\\ :\text{Angle between magnetic field and current},\mathrm{\theta }=\text{45}\mathrm{°}\\ \text{Force acting\hspace{0.17em}on the wire\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}by},\\ \text{F}={\text{BIl}}_{\text{1}}\text{sin}\mathrm{\theta }\\ =\text{BIl}\\ =1.5\times 7\times 0.2=2.1\mathrm{N}\\ \therefore \text{A force of 2}.\text{1 N acts on the wire\hspace{0.17em}in\hspace{0.17em}the\hspace{0.17em}vertically}\\ \text{downward direction\hspace{0.17em}}.\\ \left(\text{c}\right)\text{The wire in\hspace{0.17em}the\hspace{0.17em}N-S\hspace{0.17em}direction\hspace{0.17em}is lowered from the}\\ \text{axis by distance},\text{d}=\text{6}.0\text{cm}\\ \mathrm{Suppose}\mathrm{the}\mathrm{wire}\mathrm{is}\mathrm{passing}\mathrm{normally}\mathrm{to}\mathrm{the}\mathrm{axis}\mathrm{of}\\ \mathrm{cylinderical}\mathrm{magnetic}\mathrm{field},\mathrm{then}\mathrm{lowering}6\mathrm{cm}\mathrm{means}\\ \mathrm{displacing}\mathrm{the}\mathrm{wire}6\mathrm{cm}\mathrm{from}\mathrm{its}\mathrm{initial}\mathrm{position}\mathrm{towards}\\ \mathrm{to}\mathrm{the}\mathrm{end}\mathrm{of}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{cross}-\mathrm{section}.\end{array}$

$\begin{array}{l}\mathrm{From}\mathrm{the}\mathrm{given}\mathrm{figure},\mathrm{we}\mathrm{have}\\ \mathrm{x}=\sqrt{{10}^2-6^2}=8\mathrm{cm}\\ \therefore \mathrm{Length}\mathrm{of}\mathrm{wire}\mathrm{in}\mathrm{the}\mathrm{magnetic}\mathrm{field},{\mathrm{l}}_2=2\mathrm{x}=2\times 8\\ =16\mathrm{cm}\text{= 0.16 m}\\ \mathrm{Force}\mathrm{on}\mathrm{wire},\mathrm{F}={\mathrm{il}}_2\mathrm{Bsin}{90}^0\\ \mathrm{F}=7\times 0.16\times 1.5=1.68\mathrm{N}\\ \mathrm{According}\mathrm{to}\mathrm{screw}\mathrm{rule},\mathrm{the}\mathrm{direction}\mathrm{of}\mathrm{this}\mathrm{force}\mathrm{is}\\ \mathrm{vertically}\mathrm{downwards}.\\ \therefore \text{A force of 1}.\text{68 N acts\hspace{0.17em}on the wire in the\hspace{0.17em}vertically}\\ \text{downward direction}.\end{array}$

Q.6 A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?

Ans.

\begin{array}{l}\text{Here,}l\text{ength of wire},\text{l}=\text{3 cm}=0.0\text{3 m}\\ \text{Current flowing through the wire},\text{I}=\text{1}0\text{A}\\ \text{Magnitude}\text{of}m\text{agnetic field},\text{B}=0.\text{27 T}\\ \text{Angle between current and magnetic field},\theta =\text{9}0°\\ \text{Magnetic force exerted on the wire is given by}\text{the}\\ \text{relation,}\\ \text{F}=\text{BIlsin}\theta =0.\text{27}\times \text{1}0\times 0.0\text{3 sin9}0°\\ =\text{8}.\text{1}\times \text{1}{0}^{–\text{2}}\text{N}\\ \therefore \text{The magnetic force on the wire is 8}.\text{1}\times \text{1}{0}^{–\text{2}}\text{N}.\end{array}

Q.7 Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.

Ans.

\begin{array}{l}\text{Current flowing through wire A},{\text{I}}_{\text{A}}=\text{8}.0\text{A}\\ \text{Current flowing through wire B},{\text{I}}_{\text{B}}=\text{5}.0\text{A}\\ \text{Distance between the wires},\text{r}=\text{4}.0\text{cm}=0.0\text{4 m}\\ \text{Length of the section of wire A},\text{l}=\text{1}0\text{cm}=0.\text{1 m}\\ \text{Force exerted on length l due to the magnetic field is}\\ \text{given by}\text{the}\text{relation,}\\ \text{B=}\frac{{\mu }_{0}2I_A{I}_{B}l}{4\pi r}\\ \text{Here},{\mu }_0=\text{Permeability of free space}\\ =\text{4}\pi \times \text{1}{0}^{–\text{7}}{\text{TmA}}^{–\text{1}}\\ \therefore \text{B=}\frac{\text{4}\pi \times \text{1}{0}^{–\text{7}}\times 2\times 8\times 5\times 0.1}{\text{4}\pi \times 0.04}\\ =\text{2}\times \text{1}{0}^{–\text{5}}\text{N}\\ \text{This is an attractive force normal to A towards B}\text{.This}\\ \text{is}\text{because, the direction of current in the both}\text{of}the\\ \text{wires is same}.\end{array}

Q.8 A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B → inside the solenoid near its centre.

Ans.

\begin{array}{l}\text{Here,}l\text{ength of solenoid},\text{l}=\text{8}0\text{cm}=0.\text{8 m}\\ \text{Since}t\text{here are}\text{5 layers of windings of 4}00\text{turns}\\ \text{each on the solenoid,}\\ \therefore \text{Total number of turns}\text{on}the\text{solenoid},\text{N}=\text{5}\times \text{4}00\\ =\text{2}000\\ \text{Diameter of solenoid},\text{D}=\text{1}.\text{8 cm}=0.0\text{18 m}\\ \text{Current flowing}\text{through the solenoid},\text{I}=\text{8}.0\text{A}\\ \text{Magnitude of magnetic field inside the solenoid near}\\ \text{its centre is given as:}\\ \text{B=}\frac{{\mu }_{0}NI}{l}\\ \text{Here},{\mu }_0=\text{Permeability of free space}\\ =\text{4}\pi \times \text{1}{0}^{–\text{7}}{\text{TmA}}^{–\text{1}}\\ \text{B=}\frac{\text{4}\pi \times \text{1}{0}^{–\text{7}}\times 2000\times 8}{0.8}=8\pi \times \text{1}{0}^{–3}\\ =2.512\times \text{1}{0}^{–2}T\\ \therefore \text{The magnitude of the magnetic field inside the}\\ \text{solenoid near its centre is 2}.\text{512}\times \text{1}{0}^{–\text{2}}\text{T}.\end{array}

Q.9 A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30^o with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

Ans.

\begin{array}{l}\text{Here,}l\text{ength of one side of the square coil},\text{l}=\text{1}0\text{cm}\\ =0.\text{1 m}\\ \text{Current flowing through the coil},\text{I}=\text{12 A}\\ \text{Number of turns on the coil},\text{n}=\text{2}0\\ \text{Angle made by the plane of coil with uniform}\text{magnetic}\\ \text{field},\theta =\text{3}0°\\ \text{Magnitude of magnetic field},\text{B}=0.\text{8}0\text{T}\\ \text{Magnitude of the magnetic torque experienced by the}\\ \text{coil in the magnetic field is given}as:\\ \tau =\text{nBIAsin}\theta \\ \text{Area of square coil,}\text{A=l}\times \text{l}=0.\text{1}\times 0.\text{1}=0.0{\text{1 m}}^{\text{2}}\\ \therefore \tau =\text{2}0\times 0.\text{8}\times \text{12}\times 0.0\text{1}\times \text{sin3}{0}^o=0.\text{96 Nm}\\ \therefore T\text{he magnitude of torque experienced by the coil}\text{is}\\ 0.\text{96 Nm}.\end{array}

Q.10 Two moving coil meters, M₁ and M₂ have the following particulars:
R₁ = 10 Ω, N₁ = 30,
A₁ = 3.6 × 10^–3 m², B₁ = 0.25 T
R₂ = 14 Ω, N₂ = 42,
A₂ = 1.8 × 10^–3 m², B₂ = 0.50 T
(The spring constants are identical for the two meters).
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M₂ and M₁.

Ans.

\begin{array}{l}\\ \text{In}\text{case}{\text{of coil meter M}}_{\text{1}}:\\ \text{Resistance},{\text{R}}_{\text{1}}=\text{1}0\Omega ;\text{Number of turns},{\text{N}}_{\text{1}}=\text{3}0\\ C\text{ross}-\text{sectional}\text{area},{\text{A}}_{\text{1}}=\text{3}.\text{6}\times \text{1}{0}^{–\text{3}}{\text{m}}^{\text{2}}\\ \text{Strength}\text{of}m\text{agnetic field},{\text{B}}_{\text{1}}=0.\text{25 T}\\ {\text{Spring constant K}}_{\text{1}}=\text{K}\\ \text{In}\text{case}\text{of}coil{\text{meter M}}_{\text{2}}:\\ \text{Resistance},{\text{R}}_{\text{2}}=\text{14}\Omega \\ \text{Number of turns},{\text{N}}_{\text{2}}=\text{42}\\ \text{Cross}-\text{sectional}\text{area},{\text{A}}_{\text{2}}=\text{1}.\text{8}\times \text{1}{0}^{–\text{3}}{\text{m}}^{\text{2}}\\ \text{Strength}\text{of}m\text{agnetic field},{\text{B}}_{\text{2}}=0.\text{5}0\text{T}\\ \text{Spring constant},{\text{K}}_{\text{2}}=\text{K}\\ \left(\text{a}\right)\text{Current sensitivity of coil meter}{\text{M}}_{\text{1}}\text{is given as}:\\ I_{s1}=\frac{{\text{N}}_{\text{1}}{\text{B}}_{\text{1}}{\text{A}}_{\text{1}}}{{\text{K}}_{\text{1}}}\\ \text{Current sensitivity of coil meter}{\text{M}}_{\text{2}}\text{is given as}:\\ I_{s2}=\frac{{\text{N}}_2{\text{B}}_2{\text{A}}_2}{{\text{K}}_2}\\ \therefore \frac{I_{s2}}{I_{s1}}=\frac{{\text{N}}_2{\text{B}}_2{\text{A}}_2{\text{K}}_{\text{1}}}{{\text{N}}_{\text{1}}{\text{B}}_{\text{1}}{\text{A}}_{\text{1}}{\text{K}}_2}=\frac{\text{42}\times 0.\text{5}0\times \text{1}.\text{8}\times \text{1}{0}^{–\text{3}}\times K}{K\times 30\times 0.25\times 3.6\times {10}^{-3}}=1.4\\ \therefore {\text{The ratio of current sensitivity of M}}_{\text{2}}{\text{to M}}_{\text{1}}\text{is 1}.\text{4}.\\ \left(\text{b}\right)\text{Voltage sensitivity for meter}{\text{M}}_{\text{2}}\text{is given as}:\\ V_{s2}=\frac{{\text{N}}_2{\text{B}}_2{\text{A}}_2}{{\text{K}}_2{R}_2}\\ \text{Voltage sensitivity for}{\text{meter M}}_{\text{1}}\text{is given as}:\\ V_{s1}=\frac{{\text{N}}_1{\text{B}}_1{\text{A}}_1}{{\text{K}}_1{R}_1}\\ \therefore \frac{V_{s2}}{V_{s1}}=\frac{{\text{N}}_2{\text{B}}_2{\text{A}}_2{\text{K}}_1{R}_1}{{\text{N}}_1{\text{B}}_1{\text{A}}_1{\text{K}}_2{R}_2}=\frac{\text{42}\times 0.\text{5}0\times \text{1}.\text{8}\times \text{1}{0}^{–\text{3}}\times 10\times K}{K\times 14\times 30\times 0.25\times 3.6\times {10}^{-3}}=1\\ \therefore {\text{The ratio of voltage sensitivity of M}}_{\text{2}}{\text{to M}}_{\text{1}}\text{is 1}.\end{array}

Q.11 In a chamber, a uniform magnetic field of 6.5 G (1 G = 10^–4 T) is maintained. An electron is shot into the field with a speed of 4.8 × 10⁶ m s^–1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e =1.6 × 10^–19 C, m_e= 9.1×10^–31 kg)

Ans.

\begin{array}{l}\text{Here,}\text{strength}\text{of}m\text{agnetic field},\text{B}=\text{6}.\text{5 G}\\ =\text{6}.\text{5}\times \text{1}{0}^{–\text{4}}\text{T}\\ \text{Speed of electron},\text{v}=\text{4}.\text{8}\times \text{1}{0}^{\text{6}}{\text{ms}}^{\text{-1}}\\ \text{Charge on electron},\text{e}=\text{1}.\text{6}\times \text{1}{0}^{–\text{19}}\text{C}\\ \text{Mass of electron},{\text{m}}_{\text{e}}=\text{9}.\text{1}\times \text{1}{0}^{–\text{31}}\text{kg}\\ \text{Angle between the shot electron and magnetic field},\theta \\ =\text{9}0°\\ \text{Magnetic force exerted on the electron in the magnetic}\\ \text{field is given by}\text{the}\text{relation,}\\ \text{F}=\text{evB sin}\theta \\ \text{This force provides centripetal force to the moving}\\ \text{electron}\text{and}\text{the electron starts}\text{moving in a circular}\\ \text{path of radius r}.\\ \therefore \text{Centripetal force exerted on the electron}\text{is}\text{given}\text{by},\\ {\text{F}}_{\text{c}}\text{=}\frac{m{v}^2}{r}\\ \text{At equilibrium},\text{the centripetal force exerted on the}\\ \text{electron is equal to the magnetic}\text{force i}.\text{e}.,\\ {\text{F}}_{\text{c}}\text{=}F\\ \frac{m{v}^2}{r}=\text{evB sin}\theta \\ r=\frac{mv}{\text{Besin}\theta }\\ =\frac{\text{9}.\text{1}\times \text{1}{0}^{–\text{31}}\times \text{4}.\text{8}\times \text{1}{0}^{\text{6}}}{\text{6}.\text{5}\times \text{1}{0}^{–\text{4}}\times \text{1}.\text{6}\times \text{1}{0}^{–\text{19}}\times \text{sin}{90}^0}\\ =4.2x{10}^{-2}m\\ \therefore \text{The radius of the circular orbit of the electron is}\\ \text{4}.\text{2 cm}.\end{array}

Q.12 In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

Ans.

\begin{array}{l}\text{Here,}\text{strength}\text{of}m\text{agnetic field},\text{B}=\text{6}.\text{5}\times \text{1}{0}^{-\text{4}}\text{T}\\ \text{Charge on}\text{electron},\text{e}=\text{1}.\text{6}\times \text{1}{0}^{-\text{19}}\text{C}\\ \text{Mass of electron},{\text{m}}_{\text{e}}=\text{9}.\text{1}\times \text{1}{0}^{-\text{31}}\text{kg}\\ \text{Velocity of electron},\text{v}=\text{4}.\text{8}\times \text{1}{0}^{\text{6}}{\text{ms}}^{\text{-1}}\\ \text{Radius of orbit},\text{r}=\text{4}.\text{2 cm}=0.0\text{42 m}\\ \text{Frequency of revolution of electron}=\nu \\ \text{Angular frequency of electron}=\omega =\text{2}\pi \nu \\ \text{The}\text{relation}\text{connecting}v\text{elocity of electron to angular}\\ \text{frequency}\text{is}\text{given}as:\\ \text{v}=\text{r}\omega \\ \text{In the circular orbit},\text{the magnetic force on an}\text{electron}\\ \text{is balanced by the centripetal}\text{force}.\\ \therefore evB=\frac{m{v}^2}{r}\\ eB=\frac{m}{r}\left(\text{r}\omega \right)=\frac{m}{r}\left(\text{r2}\pi \nu \right)\\ \therefore \nu =\frac{Be}{2\pi m}=\frac{\text{6}.\text{5}\times \text{1}{0}^{-\text{4}}\times \text{1}.\text{6}\times \text{1}{0}^{-\text{19}}}{2\times 3.14\times \text{9}.\text{1}\times \text{1}{0}^{-\text{31}}}=18.2\times \text{1}{0}^{\text{6}}Hz\\ \therefore \text{The frequency of the electron is around 18 MHz}\\ \text{and is independent of the speed}\text{of the electron}.\end{array}

Q.13 (a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60^o with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.
(b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)

Ans.

\begin{array}{l}\left(\text{a}\right)\text{Here,}n\text{umber of turns on the circular coil},\text{n}=\text{3}0\\ \text{Radius of coil},\text{r}=\text{8}.0\text{cm}=0.0\text{8 m}\\ \text{Area of coil=}\pi {\text{r}}^{\text{2}}=\pi {\left(0.0\text{8}\right)}^2=0.0201m^2\\ \text{Current flowing through the coil},\text{I}=\text{6}.0\text{A}\\ \text{Strength}\text{of}m\text{agnetic field},\text{B}=\text{1 T}\\ \text{Angle between the magnetic}\text{field lines and normal}\\ \text{with the coil surface},\theta =\text{6}0°\\ \text{The coil experiences a torque in the magnetic field}\\ \text{and it turns}.\text{The counter torque exerted to prevent}\\ \text{the coil from turning is given as:}\\ \tau =\text{n IBA sin}\theta \to \left(\text{i}\right)\\ =\text{3}0\times \text{6}\times \text{1}\times 0.0\text{2}0\text{1}\times \text{sin6}0°=\text{3}.\text{133 Nm}\\ \left(\text{b}\right)\text{From relation (i),}\text{it can be concluded that the}\\ \text{magnitude of the applied torque is not dependent on}\\ \text{the shape of the coil}.\text{However,}i\text{t depends on the area}\\ \text{of the coil}.\\ \therefore \text{The}torquewillremain\text{unchanged if the circular coil}\\ \text{in the above case is replaced by a planar coil of some}\\ \text{irregular shape that encloses the same area}.\end{array}

Q.14 Two concentric circular coils X and Y of radii 16 cm and 10 cm, respectively, lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16 A; coil Y has 25 turns and carries a current of 18 A. The sense of the current in X is anticlockwise, and clockwise in Y, for an observer looking at the coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their centre.

Ans.

\begin{array}{l}ForcoilX:\\ \text{Radius},{\text{r}}_{\text{1}}=\text{16 cm}=0.\text{16 m}\\ \text{Number of turns,}{\text{n}}_{\text{1}}=\text{2}0\\ \text{Current},{\text{I}}_{\text{1}}=\text{16 A}\\ ForcoilY:\\ \text{Radius},{\text{r}}_{\text{2}}=\text{1}0\text{cm}=0.\text{1 m}\\ \text{Number of turns},{\text{n}}_{\text{2}}=\text{25}\\ \text{Current},{\text{I}}_{\text{2}}=\text{18 A}\\ \text{Magnetic field due to coil X at their centre is given as:}\\ {\text{B}}_{\text{1}}\text{=}\frac{{\mu }_0{\text{n}}_{\text{1}}{\text{I}}_{\text{1}}}{2{\text{r}}_{\text{1}}}\\ \text{Here},{\mu }_0=\text{Permeability of free space}\\ =4\pi \times {10}^{-7}Tm{A}^{-1}\\ \therefore {\text{B}}_{\text{1}}\text{=}\frac{4\pi \times {10}^{-7}\times 20\times 16}{2\times 0.16}=4\pi \times {10}^{-4}T(towardsEast)\\ \text{Magnetic field due to coil Y at their centre is given by},\\ {\text{B}}_2\text{=}\frac{{\mu }_0{\text{n}}_2{\text{I}}_2}{2{\text{r}}_2}=\frac{4\pi \times {10}^{-7}\times 25\times 18}{2\times 0.10}\\ =9\pi \times {10}^{-4}T(towardsWest)\\ \therefore \text{Net magnetic field,}B=B_2-B_1\\ =9\pi \times {10}^{-4}-4\pi \times {10}^{-4}\\ =5\pi \times {10}^{-4}T\\ =1.57\times {10}^{-3}T(towardsWest)\end{array}

Q.15 A magnetic field of 100 G (1 G = 10⁻⁴ T) is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about 10⁻³ m². The maximum current carrying capacity of a given coil of wire is 15 A and the number of turns per unit length that can be wound round a core is at most 1000 turns m⁻¹. Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic

Ans.

\begin{array}{l}\text{Here,}\text{strength}\text{of}m\text{agnetic field},\text{B}=\text{1}00\text{G}\\ =\text{1}00\times \text{1}{0}^{-\text{4}}\text{T}\\ \text{Number of turns per unit length},\text{n}=\text{1}000\text{turns}{\text{m}}^{-\text{1}}\\ \text{Current flowing through the coil},\text{I}=\text{15 A}\\ \text{Permeability of free space},{\mu }_{\text{0}}=4\pi \times {10}^{-7}Tm{A}^{-1}\\ \text{Magnetic field is given as:}\\ B={\mu }_{\text{0}}nI\\ \therefore nI=\frac{B}{{\mu }_{\text{0}}}=\frac{\text{1}00\times \text{1}{0}^{-\text{4}}}{4\pi \times {10}^{-7}}=7957.74\\ \approx 8000A{m}^{-1}\\ \text{If we}\text{take}\text{the length of the coil as 5}0\text{cm},\text{radius 4 cm},\\ \text{number of turns 4}00,\text{and current 1}0\text{A},\text{then these}\\ \text{values may give the magnetic field of same order}\text{.}\\ \text{So, these values are not unique and we may have other}\\ \text{suitable parameters too}\text{.}\end{array}

Q.16 For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distancex from its centre is given by,  B= μ 0 IR 2 N 2 ( x 2 +R 2 ) 3/2 ( a ) Show that this reduces to the familiar result for field at the centre of the coil. ( b ) Consider two parallel co-axial circular coils of equal radius R, and number of turns N,  carrying equal currents in the same direction,and separated by a distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R, and is given by, B = 0.72 μ 0 NI R , approximately. [Such an arrangement to produce a nearly uniform magnetic field over a small region is  known as Helmholtz coils.]

Ans.

\begin{array}{l}\text{Here,}\text{radius of}\text{the circular coil = R}\\ \text{Number of turns on the coil = N}\\ \text{Current flowing}\text{through the coil = I}\\ \text{Magnetic field at a point on its axis at distance x is given as:}\\ \text{B=}\frac{{\text{μ}}_{\text{0}}{\text{IR}}^{\text{2}}\text{N}}{\text{2}{\left({\text{x}}^{\text{2}}{\text{+R}}^{\text{2}}\right)}^{3/2}}\\ \text{Here,}{\text{μ}}_{\text{0}}\text{= Permeability of free space}\\ \left(\text{a}\right)\text{For magnetic field at the centre of the coil,}\text{x=0}\\ \therefore \text{B =}\frac{{\text{μ}}_{\text{0}}{\text{IR}}^{\text{2}}\text{N}}{{\text{2R}}^{\text{3}}}\text{=}\frac{{\text{μ}}_{\text{0}}\text{IN}}{\text{2R}}\\ \text{This}\text{is the familiar result for field at the centre of the coil}\text{.}\\ \left(\text{b}\right)\text{Radii of two parallel co-axial circular coils = R}\\ \text{Number of turns on each coil = N}\\ \text{Current flowing}\text{through both coils = I}\\ \text{Separation between both the coils = R}\\ \text{Consider point Q at distance d from the centre}\text{.}\\ \text{Then, one coil is at a distance of}\frac{\text{R}}{\text{2}}\text{+d from point Q}\text{.}\\ \therefore \text{Magnetic field at Q is given as:}\\ {\text{B}}_{\text{1}}\text{=}\frac{{\text{μ}}_{\text{0}}{\text{NIR}}^{\text{2}}}{\text{2}{\left[{\left(\frac{\text{R}}{\text{2}}\text{+d}\right)}^{\text{2}}{\text{+R}}^{\text{2}}\right]}^{3/2}}\\ \text{The other coil is at a distance of}\frac{\text{R}}{\text{2}}\text{-d from point Q}\text{.}\\ \therefore \text{Magnetic field due to this coil is given as:}\\ {\text{B}}_{\text{2}}\text{=}\frac{{\text{μ}}_{\text{0}}{\text{NIR}}^{\text{2}}}{\text{2}{\left[{\left(\frac{\text{R}}{\text{2}}\text{-d}\right)}^{\text{2}}{\text{+R}}^{\text{2}}\right]}^{3/2}}\\ \text{Total magnetic field,}{\text{B = B}}_{\text{1}}{\text{+ B}}_{\text{2}}\\ \text{=}\frac{{\text{μ}}_{\text{0}}{\text{IR}}^{\text{2}}}{\text{2}}\left[{\left\{{\left(\frac{\text{R}}{\text{2}}\text{-d}\right)}^{\text{2}}{\text{+R}}^{\text{2}}\right\}}^{-3/2}\text{+}{\left\{{\left(\frac{\text{R}}{\text{2}}\text{+d}\right)}^{\text{2}}{\text{+R}}^{\text{2}}\right\}}^{-3/2}\right]\end{array} \begin{array}{l}\text{=}\frac{{\text{μ}}_{\text{0}}{\text{IR}}^{\text{2}}}{\text{2}}\left[{\left(\frac{{\text{5R}}^{\text{2}}}{\text{4}}{\text{+d}}^{\text{2}}\text{-Rd}\right)}^{-3/2}\text{+}{\left(\frac{{\text{5R}}^{\text{2}}}{\text{4}}{\text{+d}}^{\text{2}}\text{+Rd}\right)}^{-3/2}\right]\\ \text{=}\frac{{\text{μ}}_{\text{0}}{\text{IR}}^{\text{2}}}{\text{2}}\text{×}{\left(\frac{{\text{5R}}^{\text{2}}}{\text{4}}\right)}^{-3/2}\left[{\left(\text{1+}\frac{\text{4}}{\text{5}}\frac{{\text{d}}^{\text{2}}}{{\text{R}}^{\text{2}}}\text{–}\frac{\text{4}}{\text{5}}\frac{\text{d}}{\text{R}}\right)}^{-3/2}\text{+}{\left(\text{1+}\frac{\text{4}}{\text{5}}\frac{{\text{d}}^{\text{2}}}{{\text{R}}^{\text{2}}}\text{+}\frac{\text{4}}{\text{5}}\frac{\text{d}}{\text{R}}\right)}^{{}^{-3/2}}\right]\\ \text{For}\text{d = R,}\text{neglecting}\frac{{\text{d}}^{\text{2}}}{{\text{R}}^{\text{2}}}\text{,}\text{we}\text{have,}\\ \approx \frac{{\text{μ}}_{\text{0}}{\text{IR}}^{\text{2}}}{\text{2}}\text{×}{\left(\frac{{\text{5R}}^{\text{2}}}{\text{4}}\right)}^{-3/2}\text{×}\left[{\left(\text{1-}\frac{\text{4}}{\text{5}}\frac{\text{d}}{\text{R}}\right)}^{{}^{-3/2}}\text{+}{\left(\text{1+}\frac{\text{4}}{\text{5}}\frac{\text{d}}{\text{R}}\right)}^{{}^{-3/2}}\right]\\ \approx \frac{{\text{μ}}_{\text{0}}{\text{IR}}^{\text{2}}\text{N}}{{\text{2R}}^{\text{3}}}\text{×}{\left(\frac{\text{4}}{\text{5}}\right)}^{3/2}\left[\text{1-}\frac{\text{6d}}{\text{5R}}\text{+1+}\frac{\text{6d}}{\text{5R}}\right]\\ \text{B =}{\left(\frac{\text{4}}{\text{5}}\right)}^{\text{3/2}}\frac{{\text{μ}}_{\text{0}}\text{IN}}{\text{R}}\text{= 0}\text{.72}\left(\frac{{\text{μ}}_{\text{0}}\text{IN}}{\text{R}}\right)\\ \therefore \text{It is proved that the}\text{magnetic field on the axis around the mid-point between the coils is}\text{uniform}\text{.}\end{array}

Q. 17 A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field
(a) outside the toroid,
(b) inside the core of the toroid, and
(c) in the empty space surrounded by the toroid.

Ans.

Inner radius of toroid, r₁ = 25 cm = 0.25 m

\begin{array}{l}\text{Outer radius of toroid}\text{core},{\text{r}}_{\text{2}}=\text{26 cm}=0.\text{26 m}\\ \text{Number of turns on the coil},\text{N}=\text{35}00\\ \text{Current flowing}\text{through the coil},\text{I}=\text{11 A}\\ \left(\text{a}\right)\text{Magnetic field outside the toroid is zero}.\\ \left(\text{b}\right)\text{Magnetic field inside the core of a toroid is given}\\ \text{by},\text{B}=\frac{{\mu }_{0}NI}{l}\\ \text{Here},{\mu }_0=\text{Permeability of free space}\\ =4\pi \times {10}^{-7}Tm{A}^{-1}\\ \text{Length of toroid,}\text{l}=2\pi \left[\frac{r_1+r_2}{2}\right]=\pi \left(0.25+0.26\right)\\ =0.51\pi \\ \therefore B=\frac{4\pi \times {10}^{-7}\times 3500\times 11}{0.51\pi }\approx 3.0\times {10}^{-2}T\\ \left(\text{c}\right)\text{The}m\text{agnetic field in the empty space surrounded}\\ \text{by the toroid is zero}.\end{array}

Q.18 Answer the following questions:
(a) A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle?
(b) A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction, and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment?
(c) An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path.

Ans.

(a) Since the initial velocity of the particle is either parallel or anti-parallel to the magnetic field, therefore, it travels along a straight path without suffering any deflection in the field.

(b) The final speed of the charged particle will be equal to its initial speed. This is because the magnetic force acting on the charged particle which is perpendicular to direction of its velocity can change the direction of velocity, but not its magnitude.

(c) When an electron travelling from West to East will enter a chamber having a uniform electrostatic field in the North-South direction, it will be deflected by the electric field towards north. The moving electron can remain undeflected only if the electric force acting on it is equal and opposite of the magnetic field.
Since the magnetic force is directed towards the South. Therefore, according to Fleming’s left hand rule, the magnetic field should be applied in a vertically downward direction to nullify the deflection.

Q.19 An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field (a) is transverse to its initial velocity, (b) makes an angle of 30^o with the initial velocity.

Ans.

\begin{array}{l}\text{Here,}\text{strength}ofm\text{agnetic field},\\ \text{B}=0.\text{15 T}\\ \text{Charge on electron},\\ \text{e}=\text{1}.\text{6}\times \text{1}{0}^{-\text{19}}\text{C}\\ \text{Mass of electron},\\ \text{m}=\text{9}.\text{1}\times \text{1}{0}^{-\text{31}}\text{kg}\\ \text{Potential difference},\\ \text{V}=\text{2}.0\text{kV}\\ =\text{2}\times \text{1}{0}^{\text{3}}\text{V}\\ \therefore \text{K}\text{.E}\text{. of the electron}=\text{eV}\\ \Rightarrow \text{eV}=\frac{1}{2}m{v}^2\\ \\ v=\sqrt{\frac{2eV}{m}}\to (i)\\ \text{Here},\text{v}=\text{velocity of electron}\\ \left(\text{a}\right)\text{Magnetic force on the electron provides the}\\ \text{required centripetal force to the electron}.\\ \therefore \text{The electron traces a circular path of radius r}.\\ \text{Magnetic force on the electron=Bev}\\ \text{Centripetal force=}\frac{m{v}^2}{r}\\ \therefore \text{Bev}=\frac{m{v}^2}{r}\\ r=\frac{mv}{\text{Be}}\to (ii)\\ \text{From equations}\left(i\right)\text{and}\left(ii\right),\text{we have}\\ r=\frac{m}{\text{Be}}{\left[\frac{2eV}{m}\right]}^{\frac{1}{2}}\\ =\frac{\text{9}.\text{1}\times \text{1}{0}^{-\text{31}}}{0.\text{15}\times \text{1}.\text{6}\times \text{1}{0}^{-\text{19}}}\times {\left(\frac{2\times \text{1}.\text{6}\times \text{1}{0}^{-\text{19}}\times 2\times {10}^3}{\text{9}.\text{1}\times \text{1}{0}^{-\text{31}}}\right)}^{\frac{1}{2}}\\ \\ =100.55\times \text{1}{0}^{-5}\\ =1.01\times \text{1}{0}^{-3}m=1mm\\ \therefore \text{The electron has a circular trajectory of radius 1}.0\\ \text{mm normal to the direction}\text{of}\text{magnetic}\text{field}.\\ \left(\text{b}\right)\text{When the magnetic}\text{field makes an angle}\left(\theta \right)\text{of 3}0°\\ \text{with initial velocity},\text{the initial velocity}\text{is}\text{given}\text{as:}\\ v_1=v\sin \theta \\ \text{From equation}\left(\text{2}\right),\text{the expression for new radius is}\\ \text{given}\text{as}:\\ r_1=\frac{m{v}_1}{Be}=\frac{mv\sin \theta }{Be}\\ =\frac{\text{9}.\text{1}\times \text{1}{0}^{-\text{31}}}{0.15\times \text{1}.\text{6}\times \text{1}{0}^{-\text{19}}}\times {\left[\frac{2\times \text{1}.\text{6}\times \text{1}{0}^{-\text{19}}\times 2\times {10}^3}{\text{9}.\text{1}\times \text{1}{0}^{-\text{31}}}\right]}^{\frac{1}{2}}\times \sin \text{3}0°\\ =0.5\times {10}^{-3}m=0.5mm\\ \therefore \text{The electron has a helical trajectory of radius}\\ 0.\text{5 mm along the direction}of\text{magnetic field}.\end{array}

Q.20 A magnetic field set up using Helmholtz coils (described in Exercise 4.16) is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through 15 kV enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is 9.0 × 10⁻⁵ V m⁻¹, make a simple guess as to what the beam contains. Why is the answer not unique?

Ans.

\begin{array}{l}\text{Here,}s\text{trength}\text{of}m\text{agnetic field},\text{B}=0.\text{75 T}\\ \text{Accelerating voltage},\text{V}=\text{15 kV}=\text{15}\times \text{1}{0}^{\text{3}}\text{V}\\ \text{Electrostatic field},\text{E}=\text{9}\times \text{1}{0}^{\text{5}}{\text{V m}}^{-\text{1}}\\ \text{Mass of electron}=\text{m}\\ \text{Charge of electron}=\text{e}\\ \text{Velocity of electron}=\text{v}\\ \text{Kinetic energy of electron}=\text{eV}\\ \Rightarrow \frac{1}{2}m{v}^2=\text{eV}\\ \therefore \frac{e}{m}=\frac{v^2}{2V}\to (i)\\ \text{As the particle remains undeflected by electric and}\\ \text{magnetic fields},\\ \therefore \text{The electric field is balancing the magnetic field}.\\ \therefore eE=evB\\ v=\frac{E}{B}\to (ii)\\ \text{Substituting equation}\left(ii\right)\text{in equation}\left(i\right),\text{we get}\\ \\ \frac{e}{m}=\frac{1}{2}\frac{{\left(\frac{E}{B}\right)}^2}{V}=\frac{E^2}{2V{B}^2}=\frac{{\left(\text{9}\times \text{1}{0}^{\text{5}}\right)}^2}{2\times 15000\times {\left(0.75\right)}^2}\\ =4.8\times \text{1}{0}^{7}Ck{g}^{-1}\\ \text{This value of specific charge}\frac{e}{m}\text{corresponds}\text{to}\\ \text{deuteron or deuterium ions}.\\ \text{However,}t\text{his is not a unique answer}.\text{The}o\text{ther}\\ {\text{possible answers are He}}^{++},{\text{Li}}^{+++},\text{etc}.\end{array}

Q. 21 A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires.
(a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?
(b) What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field same as before? (Ignore the mass of the wires.) g = 9.8 ms⁻².

Ans.

\begin{array}{l}\text{Here,}l\text{ength of rod},\text{l}=0.\text{45 m}\\ \text{Mass suspended}\text{by}\text{vertical}\text{wires},\text{m}=\text{6}0\text{g}\\ =\text{6}0\times \text{1}{0}^{-\text{3}}\text{kg}\\ \text{Acceleration due to gravity},\text{g}=\text{9}.{\text{8 ms}}^{\text{-2}}\\ \text{Current in the rod passing through the wire},\text{I}=\text{5 A}\\ \left(\text{a}\right)\text{Magnetic field strength}\text{is equal and opposite to the}\\ \text{weight}\text{of the wire i}.\text{e}.,\\ BIl=mg\\ \therefore B=\frac{mg}{Il}=\frac{\text{6}0\times \text{1}{0}^{-\text{3}}\times \text{9}.\text{8}}{5\times 0.\text{45}}=0.26T\\ \text{A horizontal magnetic field of}0.\text{26 T perpendicular to}\\ \text{the length of the conductor should be set}\text{up in order}\\ \text{to get zero tension in the wire}.\\ \text{The magnetic field must be such that}the\text{Fleming}’\text{s left}\\ \text{hand rule gives an upward magnetic force}.\\ \left(\text{b}\right)\text{If}\text{we reverse}\text{the direction of current},\text{then the}\\ \text{force due to magnetic field and}\text{the weight of the wire}\\ \text{act in the vertically downward direction}.\\ \therefore \text{Total tension in the wire}=\text{BIl}+\text{mg}\\ \text{=}0.\text{26}\times 5\times 0.\text{45}+\left(\text{6}0\times \text{1}{0}^{-\text{3}}\right)\times 9.8\\ =1.176N\end{array}

Q.22 The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm long and 1.5 cm apart? Is the force attractive or repulsive?

Ans.

\begin{array}{l}\text{Here,}c\text{urrent flowing}\text{through both wires},\text{I}=\text{3}00\text{A}\\ \text{Separation between the wires},\text{r}=\text{1}.\text{5 cm}=0.0\text{15 m}\\ \text{Length of each wire},\text{l}=\text{7}0\text{cm}=0.\text{7 m}\\ \text{Force between the two wires is given as:}\\ \text{F=}\frac{{\mu }_0{I}^2}{2\pi r}\\ \text{Here},{\mu }_0=\text{Permeability of free space}\\ =4\pi \times {10}^{-7}Tm{A}^{-1}\\ \therefore F=\frac{4\pi \times {10}^{-7}\times {\left(300\right)}^2}{2\pi \times 0.0\text{15}}=1.2N{m}^{-1}\\ \text{As the direction of current in the wires is opposite},\\ \text{a repulsive force exists}\text{between them}.\end{array}