Question

If 47.0 cm of copper wire (diameter = 1.00 mm) is formed into a circular loop and placed perpendicular to a uniform magnetic field that is increasing at a constant rate of 10.5mT/s, at what rate is thermal energy generated in the loop?

_______________ W

Answer 1

Ans;

     

Given: Length of wire L=47cm

Radius of wire r' =1mm

rate of Mag. field (dB/dt)=10.5mT/s

THERMAL ENERGY= ?

Thermal energy =the power(P=I²R) in the wire

So, Power P= I²R=(induced emf / R )² /R [ Current I = EMF/R]

P= (emf)²/R

But, emf = d/dt

P=(d/dt)² / R

=[d(B.A) / dt]² /R

=A² (dB/dt)² /R

=A² (dB/dt)² / (L/a) [since, Specific resistance =Ra/L]

Here, a - is the area of crossetion of wire.

Loop Area A=L²/4 [

P=[L²/4]² [dB/dt]² /(L/a)

P=L³(r' )²(dB/dt)² / [16]

Here, Specific resistance of Cu =1.68 10^-8 Ohm-m

On substituting, These values,   

We get Power or Thermal Energy is: 1.356110^-5 J/s