Question

Suppose the loop in Exercise 6.4 is stationary but the currentfeeding the electromagnet that produces the magnetic field isgradually reduced so that the field decreases from its initial valueof 0.3 T at the rate of 0.02 T s–1. If the cut is joined and the loophas a resistance of 1.6 W, how much power is dissipated by theloop as heat? What is the source of this power?

Answer 1

Given: The sides of the rectangular loop are 8 cm and 2 cm, the initial value of the magnetic field is 0.3 T, the rate of decrease of the magnetic field is 0.02  Ts -1 and the resistance of the loop is 1.6 Ω.

The area of the loop is given as,

A=lb

Where, the length of the rectangular loop is l and the width is b.

By substituting the given values in the above formula, we get

A=8× 10 −2 ×2× 10 −2 =16× 10 −4   m 2

The emf induced is given as,

e= dϕ dt

Where the change in the flux through the loop area is dϕ.

The induced emf can be written as,

e= d( AB ) dt =A dB dt

Where, the area of the loop is A and the rate of decrease of the magnetic field is dB dt .

By substituting the given values in the above expression, we get

e=16× 10 −4 ×0.02 =0.32× 10 −4  V

The induced current in the loop is given as,

i= e R

Where, the resistance of the loop is R.

By substituting the given values in the above formula, we get

i= 0.32× 10 −4 1.6 =2× 10 −5  A

The power dissipated is given as,

P= i 2 R

By substituting the given values in the above formula, we get

P= ( 2× 10 −5 ) 2 ×1.6 =6.4× 10 −10  W

Thus, the power dissipated in the loop in form of heat is 6.4× 10 −10  W and the source for this power is some external agent which changes the magnetic field with time and induced current.