Q11 of 18 Page 231

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?

Given: Length of rectangular wire = 8 cm

Breadth of rectangular wire = 2 cm


The area of the rectangular wire can be calculated as follows:


A = l × b


A = 8cm × 2cm = 16 cm2= 16 × 10-4 m2


Initial Magnitude of magnetic field B = 0.3T


Velocity of the loop, v =1 cms-1


The rate of decrease of the magnetic field i.e.


The e.m.f developed in the rectangular loop is given as:



dФ is the change in the flux across the loop = AB


⇒


Or


Therefore, substituting the values in above equation, we get:


e = 16×10-4 m2× 0.02 Ts-1


⇒ e = 0.32 × 10-4V


The current induced in the loop can be calculated as follows:


i = e/r


i = (0.32 × 10 - 4)/1.6


⇒ i = 2 × 10 - 5A


The Power loss or dissipation can be calculated as follows:


P = i2R


Where, R is resistance = 1.6Ω


⇒ P = (2 × 10-5A)2 × 1.6Ω


⇒ P = 6.4×10-10 W


The source of the power is an external agent. This is due to change in the magnetic field with time.


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