# Physics Class 12 NCERT Solutions: Chapter 6 Electromagnetic Induction Part 5

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Q: 12. A square loop of side with sides parallel to axes is moved with a velocity of in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of along the negative x-direction (that is it increases by. as one moves in the negative x-direction), and it is decreasing in time at the rate of. Determine the direction and magnitude of the induced current in the loop if its resistance is.

Answer:

Side of the square loop,

Area of the square loop,

Velocity of the loop,

Gradient of the magnetic field along negative x-direction,

And, rate of decrease of the magnetic field,

Resistance of the loop,

Rate of change of the magnetic flux due to the motion of the loop in a non-uniform magnetic field is given as:

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

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

∴Induced current,

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

Q: 13. It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet. A small flat search coil of area with closely wound turns, is positioned normal to the field direction, and then quickly snatched out of the field region. Equivalently, one can give it a quick turn to bring its plane parallel to the field direction). The total charge flown in the coil (measured by a ballistic galvanometer connected to coil) is. The combined resistance of the coil and the galvanometer is . Estimate the field strength of magnet.

Answer:

Area of the small flat search coil,

Number of turns on the coil,

Total charge flowing in the coil,

Total resistance of the coil and galvanometer,

Induced current in the coil,

Induced emf is given as:

Where,

Charge in flux

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

Initial flux through the coil,

Where,

Magnetic field strength

Final flux through the coil,

Integrating equation (3) on both sides, we have

But total charge, .

Hence, the field strength of the magnet is .