Question

In Fig. 28-47, a rectangular loop carrying current lies in the plane of a uniform magnetic field of magnitude 0.040 T . The loop consists of a single turn of flexible conducting wire that is wrapped around a flexible mount such that the dimensions of the rectangle can be changed. (The total length of the wire is not changed.) As edge length x is varied from approximately zero to its maximum value of approximately 4.0cm, the magnitude $\mathit{\tau }$of the torque on the loop changes. The maximum value of $\mathit{\tau }$is localid="1662889006282">4.80×10-8N.m . What is the current in the loop?

Short answer

The current in the loop is, i = 0.0030 A

Step-by-step solution

Given

The maximum torque on the loop is$\tau =4.80\times {10}^{-8}Nm$

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

Magnetic field, B = 0.040 T

Understanding the concept

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:

$\begin{array}{rcl}\tau & =& NiAB\sin \theta \\ A& =& sid{e}^2\end{array}$

Calculate the current in a loop

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

$\begin{array}{rcl}\tau & =& NiAB\sin \theta \end{array}$

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

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

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

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

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