Ncert Solution class 12 Physics chapter 6 Electromagnetic induction :
6.1 Predict the direction of induced current in the situation described by the following Fig. 2(a) to (f).
Ans : (a)
The given figure shows the South Pole of a bar magnet moving towards a closed loop. The direction of the induced current in a closed loop is given by Lenz's law. Using Lenz's rule, the direction of the induced current will be such that the face of the loop towards the approaching South Pole behaves as South Pole. Because the current induced in solenoid will oppose the approach of magnet and therefore the current at the nearer face should flow clockwise i.e., 𝑞𝑟𝑝𝑞
And (b)
iThe given figure shows the South Pole of a bar magnet moving towards the first closed loop (qrpq) and North Pole of the bar magnet moving away from the second loop (yzxy). The direction of the induced current in a closed loop is given by Lenz's law. Using Lenz's rule, the direction of the induced current in the first loop will be such that the face of the loop towards the approaching South Pole behaves as South Pole. Therefore, the direction of the induced current will be along prq. The direction of the induced current in the second loop will be such that the face of the loop towards the North Pole, which is moving away, behaves as South Pole. Therefore, the direction of the induced current will be along yzx.
(c) When the tapping key is just closed, the current flows in the coil in clockwise direction when viewed from left and hence magnetic flux linked with it starts increasing. According to Lenz’s law, the induced current in the adjoining coil should set up in a direction to oppose the growth of magnetic flux linked with the coil.It will happen so, if the direction of magnetic field produced by the induced current flows in the adjoining coil in anticlockwise direction i.e., the induced current is produced in the direction YZX.
(d) Due to change in rheostat setting (resistance decreased), the current in the coil (flowing anticlockwise when viewed from left) and hence magnetic flux linked with it, will start increasing. According to Lenz’s law, the induced current in the adjoining coil should set up in a direction to oppose the increasing in magnetic flux linked with the coil. Therefore, the induced current in the adjoining coil will flow in clockwise direction when viewed from left i.e., the induced current will flow in the direction zyx.
(e) Before releasing the tapping keg, the right end of the core is N-pole and the magnetic field linked with the coil is from left to right. As the tapping keg is released, the magnetic field linked with the coil will start decreasing. According to Lenz’s Law, the induced current in the adjoining coil should set up in a direction to oppose the decreasing magnetic field i.e., the induced current tries to increase the flux through first coil. It will happen so, if the direction of magnetic field produced by induced current in the adjoining coil is from left to right i.e., the induced current is produced in the direction xry.
(f) According to Lenz’s law the polarity of induced emf is such that the current produced due to it opposes the change in magnetic flux that produced it in the first place. The current flowing through the straight conductor produces magnetic field in the plane of the coil Therefore, no current will be induced in the coil.
6.2 Use Lenz's law to determine the direction of induced current in the situation described by Figure :
(a) a wire of irregular shape turning into a circular shape .
(b) a circular loop being deformed into a narrow straight wire.
Ans :
(a)When a wire of irregular shape turns into a circular shape, then its area increases, which in turn increases the flux through it. According to Lenz’s Law, the direction of induced current must decrease the flux through the irregular shaped wire,the direction of induced current should produce magnetic field in a direction outwards the plane of the paper. The direction of the current as per right Hand thumb Rule comes out to be along adcba. (i.e. anticlockwise).
(b) When a circular loop being deformed into a narrow straight wire, then the area of the loop decreases, which in turn decreases the flux through the loop. According to Lenz’s Law, the induced current must increase the flux through the loop, hence the direction of the induced current will be along a'd'c'b'a'.
6.3. A long solenoid with 15 turns per cm has a small loop area 2.0 cm placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s , what is the induced e.m.f in the loop while the current is changing ?
Solution :
$\left| \varepsilon \right| = \frac{d\phi_{B}}{dt} $
$\left| \varepsilon \right| =\frac{d}{dt}(BA) = A\frac{dB}{dt}$
since $B = \mu_{0}nI$
$\therefore \left| \varepsilon \right| = A\frac{d}{dt}(\mu_{0}nI) = \mu_{0}nA\frac{dI}{dt}$
Here , $n = 15/cm = 1500/m , A = 2.0 cm^{2}$
$= 2 \times 10^{-4} m^{2}$
$\frac{dI}{dt} = \frac{4-2}{0.1}$
$\frac{dI}{dt} = \frac{2}{0.1}$
$\frac{dI}{dt} = 20 A/s$
$|\varepsilon|= 4\pi . 10^{-7} . 1500 . 2 . 10^{-4} .20$
$|\varepsilon|= 7.5 \times 10^{-6}V$
6.4. A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the e.m.f developed across the cut if the velocity of the loop is 1 cm s in a direction normal to the (a) longer side , (b) shorter side of the loop ? For how long does the induced voltage last on each case ?
Solution :
(a)
$$\varepsilon = Blv$$
$$\varepsilon = 0.3 \times 8 \times 10^{-2} \times 10^{-2}$$
$$\varepsilon = 0.24 mV$$
Time for which e.m.f lasts = time during which shorter side of rectangle moves out of the field
$$T_{1}= \frac{l}{v} = \frac{2cm}{1cm/s} = 2s$$
(b) $$\left| \varepsilon \right| = Blv$$
$$\left| \varepsilon \right| = 0.3 \times 2 \times 10^{-2} \times 10^{-2}$$
$$\left| \varepsilon \right|=0.06mV$$
Time for which e.m.f lasts = time during which shorter side of rectangle moves out of the field
$$T_{2}= \frac{l}{v}$$
$$T_{2}= \frac{8cm}{1cm/s}$$
$$T_{2}= 8s$$
6.5. A 1.0 cm long metallic rod is rotated with an angular frequency of 400 rad /s about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the e.m.f. developed between the centre and the ring.
Solution:
Using $$\varepsilon = \frac{1}{2}Bl^{2}\omega$$
$$\varepsilon= \frac{0.5 \times 1 \times 1 \ \times 400}{2}$$
$$\varepsilon = 100 V$$
6.6. A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m/s , at right angles to the horizontal component of the earth's magnetic field , $0.30 \times 10^{-4}Wb m^{-2}$.
(a) What is the instantaneous value of the e.m.f. induced in the wire ?
(b) What is the direction of the e.m.f. ?
(c) Which end of the wire is at the higher electrical potential ?
Solution :
(a) $$\varepsilon = Bvl$$
$$\varepsilon = 0.30 \times 10^{-4}(5) (10)$$
$$\varepsilon=1.5 \times 10^{-3}V$$
(b) Using Fleming's Right hand rule , the direction of induced e.m.f. is from West to East.
(c) Since the rod will act as a source , the eastern end will be at higher electrical potential.
6.7. Current in a circuit falls from 5.0 A to 0.0 A in 0.1 sec. If an average emf of 200 V is induced , give an estimate of the self - inductance of the circuit.
Solution :
$$\varepsilon = -L\frac{di}{dt}$$
$$200 = -L (\frac{0-5}{0.1})$$
$$L = \frac{200\times0.1}{5}$$
$$L = 4 H$$
6.8. A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s. What is the change of flux linkage with the other coil ?
Solution :
$$\varepsilon = \frac{d\phi }{dt} = M\frac{di}{dt}$$
Therefore
$$d\phi = Mdi$$
$$d\phi = 1.5 \times (20-0)$$
$$d\phi = 30 Wb$$