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

`6.3\cdot 10^(-4) V`

Step-by-step explanation:

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

`\epsilon=-(\Delta \Phi)/(\Delta t)` (1)

where
`(\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(\Delta A)/(\Delta t)`

and the problem gives us:

`B=0.50 T` is the magnetic field

`(\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`