91Ó°ÊÓ

When a wire loop enters a magnetic field area, perpendicular to its plane. Then an induced current will flow through it, and the magnitude of the induced current is given by:

$I=\frac{\mathrm{Î\mu }}{R}$

We know that$\mathrm{Î\mu }=BLv$, therefore:

$I=\frac{BLv}{R}$

Putting in the values we get:

$$\begin{gathered} \mathrm{I}=\frac{\left(1.25\mathrm{T}\right)\left(0.705\mathrm{m}\right)\left(3.0\mathrm{m}/\mathrm{s}\right)}{12.5\mathrm{Î\copyright }} \\ =0.225\mathrm{A} \end{gathered}$$

Therefore, the current is 0.225 A, as the loop enters into the magnetic field area, according to Lenz’s law the direction of the induced magnetic field the induced current must be in the clock-wise direction.

We need to find the magnitude and the direction of the current induced in the circuit when the circuit is completely within the magnetic field, but it is still moving. The magnetic flux in this case is constant, hence the rate in zero. And so, the induced emf is also zero.

l = 0

Just like above when the circuit is moving out the magnetic field area, the magnetic field in perpendicular to the plane and an induced current will start through the circuit. The magnitude of the current so induced is:

$I=\frac{\mathrm{Î\mu }}{R}$

Here, $\mathrm{Î\mu }=BLv$, hence

$l=\frac{BLv}{R}$

Putting the values we get:

$$\begin{gathered} \mathrm{l}=\frac{\left(1.25\mathrm{T}\right)\left(0.750\mathrm{m}\right)\left(3.0\mathrm{m}/\mathrm{s}\right)}{12.5\mathrm{Î\copyright }} \\ =0.225\mathrm{A} \end{gathered}$$

As the loop exits the magnetic field area the direction of the induced current will be counter clockwise.

To sketch the graph of the current in this circuit as a function of time. We consider the clockwise current be positive and the counter clockwise current is negative. Hence the fiction is shown in the following graph: