If 450 cm of copper wirediameter 100 mmis formed into a cir

If 45.0 cm of copper wire(diameter = 1.00 mm)is formed into a circular loop and placed perpendicular to a uniform magnetic field t hat is increasing at a constant rate of 11.5 mT/a, at which rate is thermal energy generated in the loop?

Solution

As we need to determine the thermal energy dissipated in the loop, we will find out the amount of current that is produced in the copper wire as a result of changing magnetic flux.

The length of the wire = 45 cm

hence the radius of the loop = 45 / 2pi

or, the area = pi (45 / 2pi)² = 3.14 x 22.5² x 10^-4 = 1589.625 x 10^-4

Therefore the net rate of change of magnetic flux would be: AdB/dt

or Emf = 1589.625 x 10^-4 x 11.5 x 10^-3 = 18280.6875 x 10^-7 = 1.8281 x 10^-3 Volts

Also we know that the resisitivity of copper is 1.68 x 10^-8

Hence ther resistance of the given wire = 1.68 x 10^-8 x 45 x 10^-2 / 3.14 x 0.5 x 0.5 x 10^-6

or, R = 96.306 x 10^-4 Ohms

Therefore the rate of thermal energy generated would be same as the rate of resistive energy.

That is, E = V^2 / R = 1.8281^2 x 10^-6 / 96.306 x 10^-4 = 0.0347 x 10^-2 W