Question

A rectangular loop of sides $20cm$ and $10cm$ carries a current of $5.0A$. A uniform magnetic field of magnitude $0.20T$ exists parallel to the longer side of the loop. (a) What is the force acting on the loop? (b) What is the torque acting on the loop?

Answer 1

Given that,

Area $A=200c{m}^2$

Current $I=5.0A$

Magnetic field$B=0.20T$

For Loop ABCD

Let AB = 10 cm = CD and BC = DA = 20cm

(a). magnetic moment

$M=IA$

$M=5\times 200\times {10}^{-4}$

$M=1000\times {10}^{-4}$

$M=0.1$

Now, force of AB

$F_{AB}=I(\overrightarrow{AB}\times \overrightarrow{B})$

$F_{AB}=5\times 0.1\times 0.2$

$F_{AB}=0.1N$

Now, force of CD

$F_{CD}=-I(\overrightarrow{CD}\times \overrightarrow{B})$

$F_{CD}=-5\times 0.1\times 0.2$

$F_{CD}=-0.1N$

Now, force of BC

$F_{BC}=I(\overrightarrow{BC}\times \overrightarrow{B})$

$F_{BC}=0$

Similarly

$F_{DA}=0$

Now, the force acting on the loop

$F_{loop}=0.1-0.1+0+0$

$F_{loop}=0$

(b).The torque on the loop

$\tau =\overrightarrow{M}\times \overrightarrow{B}$

$\tau =0.1\times 0.2$

$\tau =0.02Nm$

Hence, the force on loop is 0 and the torque on the loop is $0.02Nm$