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**Q.1. A 800 turn coil of effective area 0.05 m ^{2} is kept perpendicular to a magnetic field 5 Ã— 10^{â€“5} T. When the plane of the coil is rotated by 90Â° around any of its coplanar axis in 0.1 s, the emf induced in the coil will be: (2019)**D

A: 2 V

B: 0.2 V

C: 2 x 10^{-3} V

D: 0.02 V

Ans:

Solution:

Magnetic field B = 5 Ã— 10

Number of turns in coil N = 800

Area of coil A = 0.05 m

Time taken to rotate Î”t = 0.1 s

Initial angle Î¸

Final angle Î¸

Change in magnetic flux Î”Ï†

= NBAcos90Â° â€“ BAcos0Â°

= â€“ NBA

= â€“ 800 Ã— 5 Ã— 10

= â€“ 2 Ã— 10

**Q.2. ****In which of the following devices, the eddy current effect is not used? (2019)A: Induction furnaceB: Magnetic braking in trainC: ElectromagnetD: Electric heaterAns: DSolution:**

Electric heater does not involve Eddy currents. It uses Joule's heating effect.

A: 7.14 A

B: 5.98 A

C: 14.76 A

D: 11.32 A

Ans:

Solution:

**Q.4. ****A thin diamagnetic rod is placed vertically between the poles of an electro-magnet. When the current in the electro-magnet is switched on, then the diamagnetic rod is pushed up, out of the horizontal magnetic field. Hence the rod gains gravitational potential energy. The work required to do this comes from (2018)A: the current sourceB: the magnetic fieldC: the lattice structure of the material of the rodD: the induced electric field due to the changing magnetic fieldAns: **A

Solution:

When current source is switched on, magnetic field sets up between poles on electromagnet. Diamagnetic material, due to its tendency to move from stronger to weaker field, is thus repelled out.

A: 16 Î¼C

B: 32 Î¼C

C: 16 Ï€ Î¼C

**Q.6.** **A long solenoid has 1000 turns. When a current of 4A flows through it, the magnetic flux linked with each turn of the solenoid is 4 x 10 ^{-3} Wb. The self-inductance of the solenoid is (2016)A: 1HB: 4HC: 3HD: 2HAns: **A

Solution:

Given,

Number of turns in the solenoid, N = 1000

Current, I = 4 A

Magnetic flux, straight phi subscript straight B = 4 x 10-3 Wb

Self-inductance of solenoid is given by,

Putting the value of equation in (i), we get

= 1H

A:

B:

C:

D:

Ans: A

Solution:

See figure alongside.

Let x be the distance of the centre of the frame from the long straight wire carrying current I.

Consider the point P at a distance y from the long straight wire carrying current I.

Strength of magnetic induction at point P is given by

Integrating over y from y = (x - a/2) to y = (x + a/2)

We get

**Q.8. ****A thin semicircular conducting ring (PQR) of radius â€˜râ€™ is falling with its plane vertical in a horizontal magnetic field B, as shown in figure. The potential difference developed across the ring when its speed is v, is : (2014)**

**A: Ï€rBv and R is at higher potentialB: 2rBv and R is at higher potentialC: ZeroD: BvÏ€r2/2 & P is at higher potentialAn: **B

Solution:

Effective length = 2r

Potential developed = 2rBv

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