Q: Q. 6.17. A line charge $\lambda$ per unit length is lodged uniformly onto the rim of a wheel of mass $M$ and radius $R$. The wheel has light non-conducting spokes and is free to rotate without friction about its axis (Fig. 6.22). A uniform magnetic field extends over a circular region within the rim. It is given by, $\mathrm{B}=-\mathrm{B}_{0} \mathrm{k}(\mathrm{r} \leq \mathrm{a} ; \mathrm{a}

Line charge per unit length $=\lambda=\frac{\text { Total charge }}{\text { Length }}=\frac{Q}{2 \pi r}$ Where, $r=$ Distance of the point within the wheel Mass of the wheel $=M$ Radius of the wheel $=R$ Magnetic field, $\vec{B}=-B_{0} \hat{k}$ At distance $r$, the magnetic force is balanced by the centripetal force i.e., $B Q v=\frac{M v^{2}}{r}$ Where, $v=$ linear velocity of the wheel $\therefore B 2 \pi r \lambda=\frac{M v}{r}$ $v=\frac{B 2 \pi \lambda r^{2}}{M}$ $\therefore$ Angular velocity, $\omega=\frac{v}{R}=\frac{B 2 \pi \lambda r^{2}}{M R}$ For $r \leq$ and $a<R$, we get: $\omega=-\frac{2 \boldsymbol{k}_{0} a^{2} \lambda}{M R} \hat{k}$

Question 1

Q. 6.11. Suppose the loop in Exercise 6.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3 T at the rate of $0.02 \mathrm{~T} \mathrm{~s}^{-1}$. If the cut is joined and the loop has a resistance of $1.6 \Omega$ how much power is dissipated by the loop as heat? What is the source

Accepted Answer

Sides of the rectangular loop are 8 cm and 2 cm .

Hence, area of the rectangular wire loop,

A $=$ length $	imes$ width

$=8 	imes 2=16 \mathrm{~cm}^{2}$

$=16 	imes 10^{-4} \mathrm{~m}^{2}$

Initial value of the magnetic field, $\quad B^{\prime}=0.3 \mathrm{~T}$

Rate of decrease of the magnetic field, $\frac{d B}{d t}=0.02 \mathrm{~T} / \mathrm{s}$

Emf developed in the loop is given as:

$e=\frac{d \phi}{d t}$

Where,

$d \phi=$ Change in flux through the loop area

$=\mathrm{AB}$

$$

\begin{aligned}

herefore e & =\frac{d(A B)}{d t}=\frac{A d B}{d t} \

& =16 	imes 10^{-4} 	imes 0.02=0.32 	imes 10^{-4} \mathrm{~V}

\end{aligned}

$$

Resistance of the loop, $\mathrm{R}=1.6 \Omega$

The current induced in the loop is given as:

$$

\begin{aligned}

i & =\frac{e}{R} \

& =\frac{0.32 	imes 10^{-4}}{1.6}=2 	imes 10^{-5} \mathrm{~A}

\end{aligned}

$$

Power dissipated in the loop in the form of heat is given as:

$$

\begin{aligned}

P & =i^{2} R \

& =\left(2 	imes 10^{-5}ight)^{2} 	imes 1.6 \

& =6.4 	imes 10^{-10} \mathrm{~W}

\end{aligned}

$$

The source of this heat loss is an external agent, which is responsible for changing the magnetic field with time.

Question 2

Q. 6.12. A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of $8 \mathrm{~cm} \mathrm{~s}^{-1}$ in the positive $x$-direction in an environment containing a magnetic field in the positive $z$-direction. The field is neither uniform in space nor constant in time. It has a gradient of $10^{-3} \mathrm{~T} \mathrm{~cm}^{-1}$ along the negative $x$-direction

Accepted Answer

Side of the square loop, $\mathrm{s}=12 \mathrm{~cm}=0.12 \mathrm{~m}$

Area of the square loop, $\mathrm{A}=0.12 	imes 0.12=0.0144 \mathrm{~m}^{2}$

Velocity of the loop, $v=8 \mathrm{~cm} / \mathrm{s}=0.08 \mathrm{~m} / \mathrm{s}$

Gradient of the magnetic field along negative x-direction,

$\frac{d B}{d x}=10^{-3} \mathrm{Tcm}^{-1}=10^{-1} \mathrm{Tm}^{-1}$

And, rate of decrease of the magnetic field,

$$

\frac{d B}{d t}=10^{-3} \mathrm{~T} \mathrm{~s}^{-1}

$$

Resistance of the loop, $R=4.5 \mathrm{~m} \Omega=4.5 	imes 10^{-3} \Omega$

Rate of change of the magnetic flux due to the motion of the loop in a non-uniform magnetic field is given as:

$$

\begin{aligned}

\frac{d \phi}{d t} & =A 	imes \frac{d B}{d x} 	imes v \

& =144 	imes 10^{-4} \mathrm{~m}^{2} 	imes 10^{-1} 	imes 0.08 \

& =11.52 	imes 10^{-5} \mathrm{Tm}^{2} \mathrm{~s}^{-1}

\end{aligned}

$$

Rate of change of the flux due to explicit time variation in field $B$ is given as:

$$

\begin{aligned}

\frac{d \phi^{\prime}}{d t} & =A 	imes \frac{d B}{d t} \

& =144 	imes 10^{-4} 	imes 10^{-3} \

& =1.44 	imes 10^{-5} \mathrm{~T} \mathrm{~m}^{2} \mathrm{~s}^{-1}

\end{aligned}

$$

Since the rate of change of the flux is the induced emf, the total induced emf in the loop can be calculated as:

$$

\begin{aligned}

& e=1.44 	imes 10^{-5}+11.52 	imes 10^{-5} \

& \quad=12.96 	imes 10^{-5} \mathrm{~V} \

& 	herefore 	ext { Induced current, } \quad i=\frac{e}{R} \

& =\frac{12.96 	imes 10^{-5}}{4.5 	imes 10^{-3}} \

& i=2.88 	imes 10^{-2} \mathrm{~A}

\end{aligned}

$$

Hence, the direction of the induced current is such that there is an increase in the flux through the loop along positive z-direction.

Question 3

Q. 6.13. It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet. A small flat search coil of area $2 \mathrm{~cm}^{2}$ with 25 closely wound turns, is positioned normal to the field direction, and then quickly snatched out of the field region. Equivalently, one can give it a quick $90^{\circ}$ turn to bring its plane parallel to the field direction). Th

Accepted Answer

Area of the small flat search coil, $\mathrm{A}=2 \mathrm{~cm}^{2}=2 	imes 10^{-4} \mathrm{~m}^{2}$

Number of turns on the coil, $N=25$

Total charge flowing in the coil, $\mathrm{Q}=7.5 \mathrm{mC}=7.5 	imes 10^{-3} \mathrm{C}$

Total resistance of the coil and galvanometer, $\mathrm{R}=0.50 \Omega$

Induced current in the coil,

$$

\begin{equation*}

I=\frac{	ext { Induced emf }(e)}{R} 	ag{1}

\end{equation*}

$$

Induced emf is given as:

$$

\begin{equation*}

e=-N \frac{d \phi}{d t} 	ag{2}

\end{equation*}

$$

Where,

$d \phi=$ Charge in flux

Combining equations (1) and (2), we get

$$

\begin{align*}

& I=-\frac{N \frac{d \phi}{d t}}{R} \

& I d t=-\frac{N}{R} d \phi 	ag{3}

\end{align*}

$$

Initial flux through the coil, $\phi_{i}=\mathrm{BA}$

Where,

$B=$ Magnetic field strength

Final flux through the coil, $\phi_{f}=0$

Integrating equation (3) on both sides, we have

$\int I d t=\frac{-N}{R} \int_{\phi_{i}}^{\phi_{j}} d \phi$

But total charge, $Q=\int I d t$.

$$

\begin{aligned}

herefore Q & =\frac{-N}{R}\left(\phi_{f}-\phi_{i}ight)=\frac{-N}{R}\left(-\phi_{i}ight)=+\frac{N \phi_{i}}{R} \

Q & =\frac{N B A}{R} \

herefore B & =\frac{Q R}{N A} \

& =\frac{7.5 	imes 10^{-3} 	imes 0.5}{25 	imes 2 	imes 10^{-4}}=0.75 \mathrm{~T}

\end{aligned}

$$

Hence, the field strength of the magnet is 0.75 T .

Question 4

Q. 6.14. Figure 6.20 shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K . Length of the rod $=15 \mathrm{~cm}, \mathrm{~B}=0.50 \mathrm{~T}$, resistance of the closed loop containing the rod $=9.0 \m

Accepted Answer

Length of the rod, $\mathrm{l}=15 \mathrm{~cm}=0.15 \mathrm{~m}$

Magnetic field strength, $B=0.50 \mathrm{~T}$

Resistance of the closed loop, $\mathrm{R}=9 \mathrm{~m} \Omega=9 	imes 10^{-3} \Omega$

(a) Induced emf $=9 \mathrm{mV}$; polarity of the induced emf is such that end P shows positive while end $Q$ shows negative ends. Speed of the rod, $v=12 \mathrm{~cm} / \mathrm{s}=0.12 \mathrm{~m} / \mathrm{s}$ Induced emf is given as: $\mathrm{e}=\mathrm{Bvl}$

$=0.5 	imes 0.12 	imes 0.15$

$=9 	imes 10^{-3} \mathrm{v}$

$=9 \mathrm{mV}$

The polarity of the induced emf is such that end P shows positive while end Q shows negative ends.

(b) Yes; when key K is closed, excess charge is maintained by the continuous flow of current.

When key K is open, there is excess charge built up at both ends of the rods.

When key K is closed, excess charge is maintained by the continuous flow of current. (c) Magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rod.

There is no net force on the electrons in rod PQ when key K is open and the rod is moving uniformly. This is because magnetic force is cancelled by the electric force set-up due to the excess charge of opposite nature at both ends of the rods.

(d) Retarding force exerted on the rod, $\mathrm{F}=\mathrm{IBI}$

Where,

$\mathrm{I}=$ Current flowing through the rod

$$

\begin{aligned}

& =\frac{e}{R}=\frac{9 	imes 10^{-3}}{9 	imes 10^{-3}}=1 \mathrm{~A} \

& 	herefore F=1 	imes 0.5 	imes 0.15 \

& =75 	imes 10^{-3} \mathrm{~N}

\end{aligned}

$$

(e) 9 mW ; no power is expended when key K is open.

Speed of the rod, $v=12 \mathrm{~cm} / \mathrm{s}=0.12 \mathrm{~m} / \mathrm{s}$

Hence, power is given as:

$$

\begin{aligned}

P & =F v \

& =75 	imes 10^{-3} 	imes 0.12 \

& =9 	imes 10^{-3} \mathrm{~W} \

& =9 \mathrm{~mW}

\end{aligned}

$$

When key K is open, no power is expended.

(f) 9 mW ; power is provided by an external agent.

Question 5

Q. 6.15. An air-cored solenoid with length 30 cm , area of cross-section $25 \mathrm{~cm}^{2}$ and number of turns 500 , carries a current of 2.5 A . The current is suddenly switched off in a brief time of $10^{-3} \mathrm{~s}$. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.

Accepted Answer

Length of the solenoid, $\mathrm{I}=30 \mathrm{~cm}=0.3 \mathrm{~m}$

Area of cross-section, $\mathrm{A}=25 \mathrm{~cm}^{2}=25 	imes 10^{-4} \mathrm{~m}^{2}$

Number of turns on the solenoid, N = 500

Current in the solenoid, $\mathrm{I}=2.5 \mathrm{~A}$

Current flows for time, $\mathrm{t}=10^{-3} \mathrm{~s}$

Average back emf, $\quad e=\frac{d \phi}{d t}$

Where,

$d \phi=$ Change in flux

$=$ NAB $\ldots$ (2)

Where,

$B=$ Magnetic field strength

$$

\begin{equation*}

=\mu_{0} \frac{N I}{l} 	ag{3}

\end{equation*}

$$

Where,

$\mu_{0}=$ Permeability of free space $=4 \pi 	imes 10^{-7} \mathrm{~T} \mathrm{~m} \mathrm{~A}^{-1}$

Using equations (2) and (3) in equation (1), we get

$e=\frac{\mu_{0} N^{2} I A}{l t}$

$=\frac{4 \pi 	imes 10^{-7} 	imes(500)^{2} 	imes 2.5 	imes 25 	imes 10^{-4}}{0.3 	imes 10^{-3}}=6.5 \mathrm{~V}$

Hence, the average back emf induced in the solenoid is 6.5 V .

Question 6

Q. 6.16. (a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig. 6.21.

(b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, $\mathrm{v}=10 \mathrm{~m} / \mathrm{s}$.

Calculate the induced emf in the loop at the instant when $x=0.2 m$. Take $\mathrm{a}=0.1 \

Accepted Answer

(a) Take a small element dy in the loop at a distance $y$ from the long straight wire (as shown in the given figure).

Magnetic flux associated with element $d y, d \phi=B d A$

Where, $\mathrm{dA}=$ Area of element $\mathrm{dy}=\mathrm{a} \mathrm{dy}$

$B=$ Magnetic field at distance $y$

$=\frac{\mu_{0} I}{2 \pi \mathrm{y}}$

$\mathrm{I}=$ Current in the wire

$$

\mu_{0}=	ext { Permeability of free space }=4 \pi 	imes 10^{-7}

$$

herefore d \phi=\frac{\mu_{0} I a}{2 \pi} \frac{d y}{y}

$$

y tends from x to $a+x$.

$$

\begin{aligned}

herefore \phi & =\frac{\mu_{0} I a}{2 \pi} \int_{x}^{a+x} \frac{d y}{y} \

& =\frac{\mu_{0} I a}{2 \pi}\left[\log _{e} yight]_{x}^{a+x} \

& =\frac{\mu_{0} I a}{2 \pi} \log _{e}\left(\frac{a+x}{x}ight)

\end{aligned}

$$

For mutual inductance $M$, the flux is given as:

$\phi=M I$

$$

\begin{aligned}

herefore M I & =\frac{\mu_{0} I a}{2 \pi} \log _{e}\left(\frac{a}{x}+1ight) \

M & =\frac{\mu_{0} a}{2 \pi} \log _{e}\left(\frac{a}{x}+1ight)

\end{aligned}

$$

(b) Emf induced in the loop, e = B'av $=\left(\frac{\mu_{0} I}{2 \pi x}ight) a v$

Given, $\mathrm{I}=$

50 A $x=$

$0.2 \mathrm{~m} \mathrm{a}=$

$0.1 \mathrm{~m} \mathrm{v}=$

$10 \mathrm{~m} / \mathrm{s}$

$e=\frac{4 \pi 	imes 10^{-7} 	imes 50 	imes 0.1 	imes 10}{2 \pi 	imes 0.2}$

$e=5 	imes 10^{-5} \mathrm{~V}$

Question 7

Q. 6.17. A line charge $\lambda$ per unit length is lodged uniformly onto the rim of a wheel of mass $M$ and radius $R$. The wheel has light non-conducting spokes and is free to rotate without friction about its axis (Fig. 6.22). A uniform magnetic field extends over a circular region within the rim. It is given by,

$\mathrm{B}=-\mathrm{B}_{0} \mathrm{k}(\mathrm{r} \leq \mathrm{a} ; \mathrm{a}

Accepted Answer

Line charge per unit length $=\lambda=\frac{	ext { Total charge }}{	ext { Length }}=\frac{Q}{2 \pi r}$

Where,

$r=$ Distance of the point within the wheel

Mass of the wheel $=M$

Radius of the wheel $=R$

Magnetic field, $\vec{B}=-B_{0} \hat{k}$

At distance $r$, the magnetic force is balanced by the centripetal force i.e., $B Q v=\frac{M v^{2}}{r}$

Where, $v=$ linear velocity of the wheel

$	herefore B 2 \pi r \lambda=\frac{M v}{r}$

$v=\frac{B 2 \pi \lambda r^{2}}{M}$

$	herefore$ Angular velocity, $\omega=\frac{v}{R}=\frac{B 2 \pi \lambda r^{2}}{M R}$

For $r \leq$ and $a<R$, we get:

$\omega=-\frac{2 \boldsymbol{k}_{0} a^{2} \lambda}{M R} \hat{k}$

6. Electrostatic Potential and Capacitance

Exercises - 1