Motional EMF Numerical Class-12 Nootan ISC Physics Solution Ch-11 Electromagnetic Induction. Step by step solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-12 Physics.
Motional EMF Numerical Class-12 Nootan ISC Physics Solution Ch-11 Electromagnetic Induction
| Board | ISC |
| Class | 12 |
| Subject | Physics |
| Book | Nootan |
| Chapter-11 | Electromagnetic Induction. |
| Topics | Numericals on Motional EMF |
| Academic Session | 2025-2026 |
Numericals on Motional EMF
Class-12 Nootan ISC Physics Solution Ch-11 Electromagnetic Induction
Que-12: A 50 cm long wire is moving in a uniform magnetic field of 0.8 N A^-1 m^-1 with a velocity of 2 ms^-1 perpendicular to both its length and the magnetic field. Calculate the induced emf in the wire. If the wire be moving parallel to the field, what will be the value of the induced emf?
Ans- e = Blv
=> 0.8 x 0.5 x 2 = 0.8 volt
when wire is moving parallel, there will be no any induction of emf.
Que-13: A 50 cm long horizontal wire is falling down with a speed of 6 ms^-1 perpendicular to a uniform magnetic field of 1.1 Wb m^-2. The magnetic field is directed from east to west. Calculate the emf induced in the wire. Which end of the wire (north or south) will be positive?
Ans- e = Blv
=> 1.1x 0.5 x 6 = 3.3 volt
Que-14: The axle of a car is 2.4 m long. If the car is going straight with a speed of 25 ms^-1 towards south and the vertical component of earth’s magnetic field at that place is 3 x 10^-5 T, then calculate the potential difference induced between the ends of the axle.
Ans- e = Blv
=> 3 x 10^-5 x 2.4 x 25 = 1.8 x 10^-3 volt
Que-15: A 0.4 m long straight conductor is moved in a magnetic field of induction 0.9 Wb m2 with a velocity of 7 ms^-1. Calculate the maximum emf induced in the conductor.
Ans- emax = Blv
=> 0.9 x 0.4 x 7 = 2.52 volt
Que-16: The two rails of a railway track insulated from each other and the ground, are connected to a milli-voltmeter. What will be the reading of the milli-voltmeter when a train travels at a speed of 180 km h^-1 along the track? Given that the vertical component of earth’s magnetic field is 0.2 x 10^-4 Wb m^-2 and the rails are separated by 1 m.
Ans- e = Blv
=> 0.2 x 10^-4 x 1 x 180 x 5/18
=> 1 x 10^-3 volt = 1 mV
Que-17: An aeroplane (the distance between the being 30 m) is landing down with a a velocity of 300 km h^-1. If, while landing, the wings of the aeroplane be east-west, find out the potential difference between the edges of the wings. If the wings be north-south, then? (Given: BH = 0.4 x 10^-4 T).
Ans- e = Blv
=> 0.4 x 10^-4 x 30 x 300 x 5/18 = 0.1 volt
when wing are north-south there will be no interaction of M.F. by wings, therefore emf = 0.
Que-18: A conducting rod PQ of length 20 cm and resistance 0.1 Ω rests on two smooth parallel rails AA’ and CC’ of negligible resistance. It can slide on the rails and the arrangement is positioned between the poles of a permanent magnet producing uniform magnetic field B = 0.4 T. The rails, the rod and the magnetic field are in three mutually perpendicular directions as shown in the figure. If the ends A and C of the rails are short. circuited, find the (i) external force required to move the rod with uniform velocity v = 10 cm s^-1 and (ii) power required to do so.
Ans- (i) emf e = Blv
=> (0.4 T) x (0.2 m) x (0.1 m/s) = 0.008 V
I = e / R
=> 0.008 V / 0.1 Ω = 0.08 A
Fm = ILB
=> (0.08 A) x (0.2 m) x (0.4 T) = 0.0064 N = 6.4 x 10^-3 N
(ii) Power P = Fext x v
P = (0.0064 N) x (0.1 m/s) = 0.00064 W = 6.4 × 10^-4 W
Therefore: (i) The external force required to move the rod with uniform velocity is 6.4 x 10^-3 N.
(ii) The power required to do so is 6.4 × 10^-4 W.
Que-19: A player whose height is 2.0 m runs towards east with a speed of 20 km h^-1. How much potential difference will be induced between his head and feet? Horizontal component of earth’s magnetic field, BH = 4.0 x 10^5 Wb m^-2. If he runs towards north, then?
Ans- e = Blv
=> 4 x 10^-5 x 2 x 20 x 5/18 = 4.4 x 10^-4 volt
If he runs towards north, no magnetic field will cut by him, therefore emf is zero
Que-20: A solenoid has 2000 turns wound over a length of 0.3 m. The area of cross-section is 1.2 x 10^-3 m^2. Around its central section a coil of 300 turns is closely wound. If an initial current of 2 A is reversed in 0.25s, find the emf induced in the coil.
Ans- e = μ0 (N1N2/l) A(di/dt)
=> 4 π x 10^-7 x (2000 x 300/0.3) . 1.2 x 10^-3 x (2-(-2)/0.25)
=> 48 x 10^-3 volt
=> 48 mV
— : End of Motional EMF Numerical Class-12 Nootan ISC Physics Solution Ch-11 Electromagnetic Induction . :–
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