Question

Two quarter circular loops of wire that carry a current of $14\text{A}$ are as shown in the figure below. Radius of the each quarter circle is $a=5\text{cm}$. Two uniform magnetic fields , $B=0.03\text{T}$ are directed in the $+x$ direction in one quarter circle and $-x$ direction in the other . Find the torque on the entire loop and the direction in which it will rotate.

• A. Zero
• B. $0.0008\text{Nm}$, Anti clock wise direction.
• C. $0.0008\text{Nm}$, clock wise direction.
• D. Data insufficient to determine the torque.

Answer 1

The correct option is A Zero

Torque acting on a current carrying loop is given by $$\overrightarrow{\tau }=\overrightarrow{\mu }\times \overrightarrow{B}$$For quarter circle present in first quadrant, Magnetic moment is given by $\overrightarrow{{\mu }_1}=I\left[\frac{\pi r^2}{4}\right]\left(\hat{i}\right){\text{Am}}^2$

From the figure, magnetic field in the quarter circle is $\overrightarrow{B_1}=0.03\left(\hat{i}\right)\text{T}$

Thus, torque acting on first quarter circle is $\overrightarrow{{\tau }_1}=I\left[\frac{\pi r^2}{4}\right]\left(\hat{i}\right)\times 0.03\left(\hat{i}\right)=\overrightarrow{0}$ (No direction)

For quarter circle present in third quadrant, Magnetic moment is given by $\overrightarrow{{\mu }_2}=I\left[\frac{\pi r^2}{4}\right](-\hat{i}){\text{Am}}^2$

From the figure, magnetic field in the quarter circle is $\overrightarrow{B_2}=0.03(-\hat{i})\text{T}$

Thus, torque acting on first quarter circle is $\overrightarrow{{\tau }_2}=I\left[\frac{\pi r^2}{4}\right](-\hat{i})\times 0.03(-\hat{i})=\overrightarrow{0}$ (No direction)

So, Net torque acting on the entire loop is zero.

Hence, option (a) is the correct answer.