Question

A loop of wire is at the edge of a region of space containing a uniform magnetic field B⃗ . The plane of the loop is perpendicular to the magnetic field. Now the loop is pulled out of this region in such a way that the area A of the coil inside the magnetic field region is decreasing at the constant rate c. That is, dAdt=−c, with c>0.

Answer 1

QUESTION:

Part A

The induced emf in the loop is measured to be $V$. What is the magnitude $B$ of the magnetic field that the loop was in?

Part B

For the case of a square loop of side length $L$ being pulled out of the magnetic field with constant speed $v$ (see the figure), what is the rate of change of area $c = -\dfrac{dA}{dt}$?

Answer:

Part A: $B = -\dfrac{V}{c}$

Part B: $c=-Lv$

Explanation:

Part A:

Faraday's law says that the induced voltage is equal to

$V =-N \dfrac{d\Phi_B}{dt}$,

which in our case(because we have only one loop) becomes

$V =- \dfrac{d (BA)}{dt}$,

and since the magnetic field is uniform (not changing),

$V =-B \dfrac{dA}{dt}.$

Now, we know that $\dfrac{dA}{dt} =c;$

therefore,

$V =-B c$

which gives us

$\boxed{B = -\dfrac{V}{c} }$

Part B:

The area of the loop can be written as

$A = Lx$,

where $x$ is the instantaneous length of the side along which the loop is moving.

Taking the derivative of both sides we get:

$\dfrac{dA}{dt} = -L\dfrac{dx}{dt}$,

and since $v =\dfrac{dx}{dt}$ we have

$c = \dfrac{dA}{dt} = -Lv$

$\boxed{c=-Lv}$

where the negative sign indicates that the area is decreasing.