Physics Class 12 NCERT Solutions: Chapter 6 Electromagnetic Induction Part 7

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Self inductance formula

Self Inductionce Formula

Q: 15. An air-cored solenoid with length, area of cross-section and number of turns, carries a current of A. The current is suddenly switched off in a brief time of. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.

Answer:

Length of the solenoid,

Area of cross-section,

Number of turns on the solenoid,

Current in the solenoid,

Current flows for time,

Average back emf,

Where,

Change in flux

Where,

Magnetic field strength

Where,

Permeability of free space

Using equations (2) and (3) in equation (1), we get

Hence, the average back emf induced in the solenoid is.

Q: 16. (A) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Figure.

(B) Now assume that the straight wire carries a current of and the loop is moved to the right with a constant velocity, .

Calculate the induced emf in the loop at the instant when .

Take and assume that the loop has a large resistance.

Ch 1 emf in the loop

Q 16 Emf in the Loop

Answer:

(A) Take a small element in the loop at a distance from the long straight wire (as shown in the given figure).

Q 16 square loop A

Q 16 Square Loop A

Magnetic flux associated with element

Where,

Area of element

Magnetic field at distance y

Permeability of free space

Tends from x to

For mutual inductance M, the flux is given as:

(B) Emf induced in the loop,

Given,

Developed by: