Chapter 6: Electromagnetic Induction

Ans: We are given,

Sides of the rectangular loop are 8cm and 2cm. Hence, area of the rectangular wire loop would be,

\[A\text{ }=\text{ }length\text{ }\times \text{ }width\]

Initial value of the magnetic field, \[B'=0.3T\]

Rate of decrease of the magnetic field, $\frac{dB}{dt}=0.02T/s$

Emf developed in the loop is given as:

$\varepsilon =\frac{d\phi }{dt}$

Where,

$\varepsilon =\frac{d\left( AB \right)}{dt}=\frac{AdB}{dt}$

$\Rightarrow \varepsilon =16\times {{10}^{-4}}\times 0.02=0.32\times {{10}^{-4}}V$

Resistance of the loop, $R=1.6\Omega $

The current induced in the loop could be given as:

$i=\frac{\varepsilon }{R}$

Substituting the given values, 

$\Rightarrow i=\frac{0.32\times {{10}^{-4}}}{1.6}=2\times {{10}^{-5}}A$

Power dissipated in the loop in the form of heat could be given as:

$P={{i}^{2}}R$

$\Rightarrow P={{\left( 2\times {{10}^{-5}} \right)}^{2}}\times 1.6$

$\therefore P=6.4\times {{10}^{-10}}W$

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

Ans: We are given, 

Side of the square loop, \[s=12cm=0.12m\]

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

Velocity of the loop, \[v=8cm/s=0.08m/s\]

Gradient of the magnetic field along negative x-direction,

$\frac{dB}{dx}={{10}^{-3}}Tc{{m}^{-1}}={{10}^{-1}}T{{m}^{-1}}$

And, rate of decrease of the magnetic field,

$\frac{dB}{dt}={{10}^{-3}}T{{s}^{-1}}$

Resistance of the loop, 

$R=4.5m\Omega =4.5\times {{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:

$\frac{d\phi }{dt}=A\times {\frac{dB}{dx}}\times v$

$\Rightarrow \frac{d\phi }{dt}=144\times {{10}^{-4}}{{m}^{2}}\times {{10}^{-1}}\times 0.08$

$\Rightarrow \frac{d\phi }{dt}=11.52\times {{10}^{-5}}T{{m}^{2}}{{s}^{-1}}$

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

$\frac{d\phi '}{dt}=A\times \frac{dB}{dx}$

$\Rightarrow \frac{d\phi '}{dt}=144\times {{10}^{-4}}\times {{10}^{-3}}=1.44\times {{10}^{-5}}T{{m}^{2}}{{s}^{-1}}$

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

\[e=1.44\times {{10}^{-5}}+11.52\times {{10}^{-5}}\]

\[=12.96\times {{10}^{-5}}V\]

\[\therefore \text{Induced current, }i=\frac{e}{R}\]

\[\Rightarrow i=\frac{12.96\times {{10}^{-5}}}{4.5\times {{10}^{-3}}}\]

\[\therefore i=2.88\times {{10}^{-2}}A\]

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

Ans: We are given the following:

Area of the small flat search coil, \[A=2c{{m}^{2}}=2\times {{10}^{-4}}{{m}^{2}}\]

Number of turns on the coil, \[N=25\]

Total charge flowing in the coil, \[Q=7.5mC=7.5\times {{10}^{-3}}C\]

Total resistance of the coil and galvanometer, \[R=0.50\Omega \]

Induced current in the coil,

\[I=\frac{\text{Induced emf}(\varepsilon )}{R}\]……………………………………………. (1)

Induced \[emf\] is given as:

\[\varepsilon =-N\frac{d\phi }{dt}\]………………………………………………… (2) 

Where, \[d\phi =\text{Induced flux}\]

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

\[I=-\frac{N\frac{d\phi }{dt}}{R}\]

\[Idt=-\frac{N}{R}d\phi \]……………………………………………. (3)

Initial flux through the coil, \[{{\phi }_{i}}=BA\]

Where,

\[B=\text{Magnetic field strength}\]

Final flux through the coil, \[{{\phi }_{f}}=0\]

Integrating equation (3) on both sides, we have

\[\int{Idt=-\frac{N}{R}\int\limits_{{{\phi }_{i}}}^{{{\phi }_{f}}}{d\phi }}\]

But total charge could be given as, \[Q=\int{Idt}\]

\[\Rightarrow Q=\frac{-N}{R}({{\phi }_{f}}-{{\phi }_{i}})=\frac{-N}{R}(-{{\phi }_{i}})=\frac{+N{{\phi }_{i}}}{R}\]

\[Q=\frac{NBA}{R}\]

\[\Rightarrow B=\frac{QR}{NA}\]

Substituting the given values, we get, 

\[\Rightarrow B=\frac{7.5\times {{10}^{-3}}\times 0.5}{25\times 2\times {{10}^{-4}}}\]

\[\therefore B=0.75T\]

Therefore, the field strength of the magnet is found to be \[0.75T\].

Ans: We are given:

Length of the rod, \[l=15cm=0.15m\]

Magnetic field strength, \[B=0.50T\]

Resistance of the closed loop, \[R=9m\Omega =9\times {{10}^{-3}}\Omega \]

Induced \[emf=9mV\]

Here, polarity of the induced \[emf\] is such that end P shows positive while end Q shows negative ends. 

Speed of the rod,\[v=12cm/s=0.12m/s\]

We know that the induced \[emf\] could be given as: \[\varepsilon =Bvl\]

Substituting the given values, we get, 

\[\varepsilon =0.5\times 0.12\times 0.15\]

\[\Rightarrow \varepsilon =9\times {{10}^{-3}}v\]

\[\therefore \varepsilon =9mV\]

Therefore, the magnitude of the induced emf is found to be \[\varepsilon =9mV\] an =d the polarity of the induced emf is such that end P shows positive while end Q shows negative.

Ans: Yes; when key K is closed, excess charge could be maintained by the continuous flow of current. When key K is open, there is excess charge built up at both rod ends but when key K is closed, excess charge is maintained by the continuous flow of current.

Ans: Magnetic force is cancelled by the electric force that is set-up due to the excess charge of opposite nature at both rod ends. There is no net force on the electrons in rod PQ when key K is open and the rod would move 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.

Ans: We know that the retarding force exerted on the rod could be given by, \[F=IBl\]

Where,

\[I=\text{Current flowing through the rod}\]

Substituting the given values, we get, 

\[I=\frac{e}{R}=\frac{9\times {{10}^{-3}}}{9\times {{10}^{-3}}}=1A\]

\[\Rightarrow F=1\times 0.5\times 0.15\]

\[\therefore F=75\times {{10}^{-3}}N\]

Therefore, we found the retarding force on the rod when the key K is closed to be, 

\[F=75\times {{10}^{-3}}N\]

Ans: We are given:

Speed of the rod, \[v=12cm/s=0.12m/s\]

Now, power could be given as:

\[P=Fv\]

Substituting the given values, we get, 

\[\Rightarrow P=75\times {{10}^{-3}}\times 0.12\]

\[\Rightarrow P=9\times {{10}^{-3}}W\]

\[\therefore P=9mW\]

Therefore, we found the power that is required (by an external agent) to keep the rod moving at the same speed \[(=12cm\cdot {{s}^{-1}})\] when K is closed to be \[P=9mW\]and when key K is open, no power is expended. 

e) How much power is dissipated as heat in the closed circuit? What is the source of this power?

Ans: We know that, 

\[\text{Power dissipated as heat , P=}{{I}^{2}}R\]

\[\Rightarrow P={{(1)}^{2}}\times 9\times {{10}^{-3}}\]

\[\therefore P=9mW\]

The power dissipated as heat in the closed circuit is found to be \[P=9mW\] and the source of this power is found to be an external agent.

f) What is the induced \[emf\] in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?

Ans: In this case, no \[emf\]would be induced in the coil because the motion of the rod does not cut across the field lines.

Ans: We are given the following: 

Length of the solenoid, \[l=30cm=0.3m\]

Area of cross-section, \[A=25c{{m}^{2}}=25\times {{10}^{-4}}{{m}^{2}}\]

Number of turns on the solenoid, \[N=500\]

Current in the solenoid, \[I=2.5A\]

Current flows for time, \[t={{10}^{-3}}s\]

Average back \[emf\], \[e=\frac{d\phi }{dt}\]……………………………………. (1)

Where,

\[d\phi =\text{Change in flux=}NAB\]…………………………………………… (2)

Where, \[B=\text{Magnetic field strength}={{\mu }_{0}}\frac{NI}{l}\]………………………… (3)

Where,\[{{\mu }_{0}}=\text{Permeability of free space}=4\pi \times {{10}^{-7}}Tm{{A}^{-1}}\]

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

\[\varepsilon =\frac{{{\mu }_{0}}{{N}^{2}}IA}{lt}\]

\[\Rightarrow \varepsilon =\frac{4\pi \times {{10}^{-7}}\times {{(500)}^{2}}\times 2.5\times 25\times {{10}^{-4}}}{0.3\times {{10}^{-3}}}\]

\[\therefore \varepsilon =6.5V\]

Hence, the average back \[emf\]induced in the solenoid is found to be \[6.5V\].

Ans: Consider 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 \[dy,d\phi =BdA\]

Where, \[dA=\text{Area of element }dy=\text{a }dy\]

\[B=\text{magnetic field at distance y =}\frac{{{\mu }_{0}}I}{2\pi y}\]

\[I=\text{Current in the wire}\]

\[{{\mu }_{0}}=\text{Permeability of free space}=4\pi \times {{10}^{-7}}H/m\]

Carrying out the substitutions accordingly, we get, 

\[\Rightarrow d\phi =\frac{{{\mu }_{0}}Ia}{2\pi }\frac{dy}{y}\]

\[\Rightarrow \phi =\frac{{{\mu }_{0}}Ia}{2\pi }\int{\frac{dy}{y}}\]

Now, the limit of y will be from x to \[a+x\], on applying the limits we get, 

\[\Rightarrow \phi =\frac{{{\mu }_{0}}Ia}{2x}\int\limits_{x}^{a+x}{\frac{dy}{y}}\]

\[\Rightarrow \phi =\frac{{{\mu }_{0}}Ia}{2\pi }\left[ {{\log }_{e}}y \right]_{x}^{a+x}\]

\[\Rightarrow \phi =\frac{{{\mu }_{0}}Ia}{2\pi }{{\log }_{e}}\left( \frac{a+x}{x} \right)\]

For mutual inductance M, the flux could be given as: 

\[\phi =MI\]

\[\Rightarrow MI=\frac{{{\mu }_{0}}Ia}{2\pi }{{\log }_{e}}\left( \frac{a}{x}+1 \right)\]

\[\therefore M=\frac{{{\mu }_{0}}a}{2\pi }{{\log }_{e}}\left( \frac{a}{x}+1 \right)\]

Therefore, the expression for the mutual inductance between the given long straight wire and the square loop of side a is found to be, 

\[M=\frac{{{\mu }_{0}}a}{2\pi }{{\log }_{e}}\left( \frac{a}{x}+1 \right)\]

Ans: We know that, the \[Emf\]induced in the loop, \[\varepsilon =B'av=\left( \frac{{{\mu }_{0}}I}{2\pi x} \right)av\]

We are given the following, 

\[I=50A\]

\[x=0.2m\]

\[a=0.1m\]

\[v=10m/s\]

On substituting the given values into the equation, we get, 

\[\varepsilon =\frac{4\pi \times {{10}^{-7}}\times 50\times 0.1\times 10}{2\pi \times 0.2}\]

\[\therefore \varepsilon =5\times {{10}^{-5}}V\]

Therefore, induced \[emf\] in the loop at the given instant is found to be, \[\varepsilon =5\times {{10}^{-5}}V\]. 

Ans: We know that, the 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, \[\vec{B}=-{{B}_{0}}k\]

At distance r, the magnetic force would be balanced by the centripetal force i.e.,

\[BQv=\frac{M{{v}^{2}}}{r}\]

Where,

\[v=\text{linear velocity of the wheel}\]

\[\Rightarrow \text{B2}\pi \text{r}\lambda =\frac{Mv}{r}\]

\[\Rightarrow v=\frac{B2\pi \lambda {{r}^{2}}}{M}\]

\[\therefore \text{Angular velocity,}\omega =\frac{v}{R}=\frac{B2\pi \lambda {{r}^{2}}}{M}\]

\[{r\leq a}\], and \[{a<R}\] we would get:

\[\omega =\frac{2\pi {{B}_{0}}{{a}^{2}}\lambda }{MR}k\]

Therefore, we found the angular velocity of the wheel after the field is suddenly switched off to be given as, 

\[\omega =\frac{2\pi {{B}_{0}}{{a}^{2}}\lambda }{MR}k\]