91Ó°ÊÓ

The maximum torque on the loop is$\mathrm{τ}=4.80×{10}^{-8}Nm$

The maximum possible length of loop L = 4.0 cm = 0.04 m

Magnetic field, B = 0.040 T

We use the equation of torque acting on a current loop. By using this equation and considering the maximum area of loop (as the maximum torque is given), we can calculate the current in a loop.

Formulae:

$\begin{array}{rcl}\mathrm{τ}& =& NiAB\sin \mathrm{θ}\\ A& =& sid{e}^2\end{array}$

The maximum possible length of a loop is 0.04 cm.

So, the length of wire will be approximately 0.08 cm.

Now, the equation for a torque on a current loop is

$\begin{array}{rcl}\mathrm{τ}& =& NiAB\sin \mathrm{θ}\end{array}$

We can see that the torque on a current loop can be maximum when area of loop A is maximum.

For the rectangular loop, the maximum area is obtained when it becomes a square.

So, if we make the 8.0 cm wire in a square shape, the side of square will be

$\begin{array}{rcl}side& =& \frac{1}{4}×0.08\\ & =& 0.02m\end{array}$

So, the area of square is

$\begin{array}{rcl}A& =& sid{e}^2\\ & =& 0.{02}^2\\ & =& 0.0004m^2\end{array}$

Now, rearranging the equation of torque for current and substituting the value of A, we get

$\begin{array}{rcl}i& =& \frac{\mathrm{τ}}{NAB\sin \mathrm{θ}}\\ i& =& \frac{4.80×{10}^{-8}}{1×0.0004×0.04×\sin \left({90}^o\right)}\\ i& =& 3.00×{10}^{-3}A\\ & =& 0.0030A\end{array}$

Hence, the current in the loop is, i = 0.0030 A