A small square loop of side a is placed at the centre of a large circular loop of radius R ( R a ). The two loops are coplanar. If an alternating current I = I₀ ( t) is passed through the large circular loop, the peak induced EMF in the small square loop is:

Step-by-step solution

The magnetic field produced by the large circular loop at its centre is: B = ₀ I 2R = ₀ I₀ ( t) 2R Since the square loop is very small ( a R ), the magnetic field can be considered uniform over its area. The magnetic flux passing through the small square loop is: = B Area = ( ₀ I₀ ( t) 2R ) a^2 = ₀ I₀ a^2 2R ( t) According to Faraday's law of induction, the induced EMF E in the small loop is: E = - d dt = - d dt ( ₀ I₀ a^2 2R ( t) ) E = - ₀ I₀ a^2 2R ( t) The peak value of the induced EMF is the amplitude of this e