Class 12 Physics
CBSE and UP Board
Electromagnetic Induction
Chapter-6 Exercise 6.11
Additional Exercise
NCERT Solutions for Class 12 Physics Chapter 6 Question 11
Suppose the loop in Exercise 6.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3 T at the rate of 0.02 T s⁻¹. If the cut is joined and the loop has a resistance of 1.6Ω how much power is dissipated by the loop as heat? What is the source of this power?
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Sides of the rectangular loop are 8 cm and 2 cm.
Hence, area of the rectangular wire loop,
A = length x width = 8 x 2 = 16 cm2 = 16 x 10⁻⁴ m2
Initial value of the magnetic field, B’ = 0.3 T
Rate of decrease of the magnetic field, dB/dt = 0.02 T/s
Emf developed in the loop is given as: e = dφ/dt
Where, dφ= Change in flux through the loop area = AB
Therefore, e = d(AB)/dt = AdB/dt
= 16 x 10⁻⁴ x 0.02 = 0.32 x 10⁻⁴ V
Resistance of the loop, R = 1.6 Ω
The current induced in the loop is given as: i = e/R
= (0.32 x 10⁻⁴)/1.6 = 2 x 10⁻⁵ A
Power dissipated in the loop in the form of heat is given as:
P = i2R
= (2 x 10⁻5)2 x 1.6 = 6.4 x 10⁻¹⁰ W
The source of this heat loss is an external agent, which is responsible for changing the magnetic field with time.