Question

A circular copper loop is placed perpendicular to a uniform magnetic field of 0.50 T. Due to external forces, the area of the loop decreases at a rate of 1.26 x 10^-3 m^2/s. Determine the induced emf in the loop.

Answer 1

Answer:

$6.3\cdot 10^{-4} V$

Explanation:

According to Faraday-Newmann-Lenz, the induced emf in the loop is given by:

$\epsilon=-\frac{\Delta \Phi}{\Delta t}$ (1)

where $\frac{\Delta \Phi}{\Delta t}$ is the rate of variation of the magnetic flux through the loop.

We know that the magnetic flux through the loop is given by

$\Phi = BA$

where B is the magnetic field and A is the area of the loop. Since the magnetic field is constant, we can write the variation of flux as

$\Delta \Phi = B \Delta A$

So eq.(1) becomes

$\epsilon=-B\frac{\Delta A}{\Delta t}$

and the problem gives us:

$B=0.50 T$ is the magnetic field

$\frac{\Delta \Phi}{\Delta t}=-1.26\cdot 10^{-3} m^2/s$ is the rate at which the area changes

Substituting into the equation, we find

$\epsilon=-(0.50 T)(-1.26\cdot 10^{-3} m^2/s)=6.3\cdot 10^{-4} V=0.63 mV$