Chapter 6: Electromagnetic Induction

Ans: The direction of the induced current in a closed loop could be given by Lenz’s law. The following pairs of figures show the direction of the induced current when the North pole of a bar magnet is moved towards and away from a closed loop respectively.

Now, by using Lenz’s rule, the direction of the induced current in the given situation is found to be along qrpq.

Ans: On using Lenz’s law, we find the direction of the induced current here to be along prqp.

Ans: Using Lenz’s law, we find the direction of the induced current to be along yzxy.

Ans: Using Lenz’s law, we find the direction of the induced current to be along zyxz.

e)

Ans: Using Lenz’s law, we found the direction of the induced current to be along xryx.

f)

Ans: Here we find that, no current is induced since the field lines are lying in the same plane as that of the closed loop.

Ans: According to Lenz's law, the direction of the induced emf is such that it tends to produce a current that would oppose the change in the magnetic flux that produced it.

The wire is expanding to form a circle, which means that force would be acting outwards on each part of the wire because of the magnetic field (acting in the downwards direction). Now, the direction of induced current should be such that it will produce magnetic field in the upward direction (towards the reader). Therefore, the force on wire will be towards the inward direction, i.e., induced current would be flowing in anticlockwise direction in the loop from cbad.

Ans: On deforming the shape of a circular loop into a narrow straight wire, the flux piercing the surface decreases. Therefore, the induced current flows along abcd according to Lenz’s law. 

Ans: We are given the following information:

Number of turns on the solenoid \[=15turns/cm=1500turns/m\]

Number of turns per unit length, \[n=1500\text{ }turns\]

The solenoid has a small loop of area,$A=2.0c{{m}^{2}}=2\times {{10}^{-4}}{{m}^{2}}$

Current carried by the solenoid changes from\[2A\text{ }to\text{ }4A\].

Now, the change in current in the solenoid, \[di=4-2=2A\]

Change in time, \[dt=0.1\text{ }s\]

Induced emf in the solenoid could be given by Faraday’s law as:

$\varepsilon =\frac{d\phi }{dt}$………………………… (1)

Where, induced flux through the small loop, $\phi =BA$…………… (2) 

Equation (1) would now reduce to:

$\varepsilon =\frac{d}{dt}\left( BA \right)=A{{\mu }_{0}}n\times \left( \frac{di}{dt} \right)$

Substituting the given values into this equation, we get, 

$\varepsilon =2\times {{10}^{-4}}\times 4\pi \times {{10}^{-7}}\times 1500\times \frac{2}{0.1}$

$\therefore \varepsilon =7.54\times {{10}^{-6}}V$

Therefore, the induced voltage in the loop is found to be, $\varepsilon =7.54\times {{10}^{-6}}V$. 

Ans: We are given the following, 

Length of the rectangular wire, \[l=8cm=0.08\text{ }m\]

Width of the rectangular wire, \[b=2cm=0.02m\]

Now, the area of the rectangular loop,

$A=lb=0.08\times 0.02=16\times {{10}^{-4}}{{m}^{2}}$

Magnetic field strength, \[B=0.3T\]

Velocity of the loop, \[v=1\text{ }cm/s=0.01\text{ }m/s\]

Emf developed in the loop could be given as:

\[\varepsilon =Blv\]

Substituting the given values, 

$\varepsilon =0.3\times 0.08\times 0.01=2.4\times {{10}^{-4}}V$

Time taken to travel along the width, $t=\frac{Distance\text{ }travelled}{\text{velocity}}=\frac{b}{v}$

$\Rightarrow t=\frac{0.02}{0.01}=2s$

Therefore, the induced voltage is found to be $2.4\times {{10}^{-4}}V$which lasts for 2s. 

Ans: We know that, Emf developed in the loop could be given as:

\[\varepsilon =Blv\]

Substituting the given values, 

$\varepsilon =0.3\times 0.02\times 0.01=0.6\times {{10}^{-4}}V$

Time taken to travel along the width, $t=\frac{Distance\text{ }travelled}{\text{velocity}}=\frac{l}{v}$

$\Rightarrow t=\frac{0.08}{0.01}=8s$

Therefore, the induced voltage is found to be $0.6\times {{10}^{-4}}V$which lasts for 8s. 

Ans: We are given the following:

Length of the rod, \[l=1m\]

Angular frequency, $\omega =400rad/s$

Magnetic field strength, \[B=0.5T\]

One end of the rod has zero linear velocity, while the other end has a linear velocity of $l\omega $. 

Average linear velocity of the rod, $\nu =\frac{l\omega +0}{2}=\frac{l\omega }{2}$

Emf developed between the centre and the ring,

$\varepsilon =Blv=Bl\left( \frac{l\omega }{2} \right)=\left( \frac{B{{l}^{2}}\omega }{2} \right)$

On substituting the given values, 

$\therefore \varepsilon =\frac{0.5\times {{\left( 1 \right)}^{2}}\times 400}{2}=100V$

Therefore, the emf developed between the centre and the ring is \[100V\].

Ans: We are given:

Max induced emf \[=0.603V\]

Average induced emf \[=0V\]

Max current in the coil \[=0.0603A\]

Average power loss \[=0.018W\text{ }\](Power comes from the external rotor)

Radius of the circular coil, \[r=8cm=0.08m\]

Area of the coil, $A=\pi {{r}^{2}}=\pi \times {{\left( 0.08 \right)}^{2}}{{m}^{2}}$

Number of turns on the coil, \[N=20\]

Angular speed, $\omega =50rad/s$

Magnetic field strength, $B=3\times {{10}^{-2}}T$

Resistance of the loop, \[R=10\text{ }\Omega \]

Maximum induced emf could be given as:

$\varepsilon =N\omega AB=20\times 50\times \pi \times {{\left( 0.08 \right)}^{2}}\times 3\times {{10}^{-2}}$

$\therefore \varepsilon =0.603V$

The maximum emf induced in the coil is found to be \[0.603V\].

Over a full cycle, the average emf induced in the coil is found to be zero.

Maximum current is given as:

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

$\Rightarrow I=\frac{0.603}{10}$

$\Rightarrow I=0.0603A$

Average power loss due to joule heating:

$\therefore P=\frac{eI}{2}=\frac{0.603\times 0.0603}{2}=0.018W$

We know that the current induced in the coil would produce a torque opposing the rotation of the coil. Since the rotor is an external agent, it must supply a torque to counter this torque in order to keep the coil rotating uniformly. Hence, dissipated power comes from the external rotor.

Ans: We are given the following:

Length of the wire, \[l=10m\]

Falling speed of the wire, \[v=5.0m/s\]

Magnetic field strength, $B=0.3\times {{10}^{-4}}Wb{{m}^{-2}}$

Emf induced in the wire is thus found to be, 

$\varepsilon =Blv$

\[\Rightarrow \varepsilon =0.3\times {{10}^{-4}}\times 5\times 10\]

$\therefore \varepsilon =1.5\times {{10}^{-3}}V$

Hence, the emf induced in the wire is $\varepsilon =1.5\times {{10}^{-3}}V$. 

Ans: Using Fleming’s rule, we find that the direction of the induced emf is from West to East.

Ans: The eastern end of the wire is the end that is at higher potential.

Ans: We are given the following:

Initial current, ${{I}_{1}}=5.0A$

Final current, ${{I}_{2}}=0.0A$

Change in current, $dI={{I}_{1}}-{{I}_{2}}=5A$

Time taken for the change, \[t=0.1s\]

Average emf, $\varepsilon =200V$

For self-inductance (L) of the coil, we have the relation for average emf that could be given as: 

$\varepsilon =L\frac{di}{dt}$

$\Rightarrow L=\frac{\varepsilon }{\left( \frac{di}{dt} \right)}$

Substituting the given values we get, 

$\therefore L=\frac{200}{\left( \frac{5}{0.1} \right)}=4H$

Therefore, we found the self induction in the coil to be 4H. 

Ans: We are given the following, 

Mutual inductance of a pair of coils, $\mu =1.5H$

Initial current, ${{I}_{1}}=0A$

Final current, ${{I}_{2}}=20A$

Change in current, $dI={{I}_{2}}-{{I}_{1}}=20-0=20A$

Time taken for the change, \[t=0.5s\]

Induced emf, $\varepsilon =\frac{d\phi }{dt}$…………………. (1)

Where, $d\phi $is the change in the flux linkage with the coil.

Emf is related with mutual inductance could be given as:

$\varepsilon =\mu \frac{dI}{dt}$…………………….. (2)

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

$\frac{d\phi }{dt}=\mu \frac{dI}{dt}$

$\Rightarrow d\phi =1.5\times \left( 20 \right)$

$\therefore d\phi =30Wb$

Hence, we found the change in the flux linkage to be 30Wb.

Ans: Speed of the jet plane, \[v=1800\text{ }km/h=500\text{ }m/s\]

Wingspan of jet plane, \[l=25m\]

Earth’s magnetic field strength, $B=5.0\times {{10}^{-4}}T$

Angle of dip, $\delta =30{}^\circ $

Vertical component of Earth’s magnetic field could be given by, 

${{B}_{V}}=B\sin \delta $

$\Rightarrow {{B}_{V}}=5\times {{10}^{-4}}\sin 30{}^\circ =2.5\times {{10}^{-4}}T$

Voltage difference between the ends of the wing can be calculated as,

$\varepsilon ={{B}_{V}}\times l\times v$

Substituting the given values, 

$\Rightarrow \varepsilon =2.5\times {{10}^{-4}}\times 25\times 500$

$\therefore \varepsilon =3.125V$

Hence, the voltage difference developed between the ends of the wings is \[3.125V\].