Physics - Magnetism Question with Solution | TestHub

The magnetic moment of the loop is $M=IA$, where $I$ is the current and $A$ is the area. For a square loop of side $L=0.1$ m and current $I=5$ A, $M=5\times (0.1{)}^2=5\times 0.01=0.05$ A m${}^2$.

The torque on the loop is given by $\tau =MB\sin {\theta }^{\prime }$, where ${\theta }^{\prime }$ is the angle between the magnetic moment vector (normal to the loop's plane) and the magnetic field. If the plane of the loop makes an angle of $30^{\circ }$ with the field, then ${\theta }^{\prime }=90^{\circ }-30^{\circ }=60^{\circ }$.

So, $\tau =(0.05{\text{ A m}}^2)\times (0.2\text{ T})\times \sin (60^{\circ })=0.01\times \frac{\sqrt{3}}{2}=0.005\sqrt{3}$ N m. Approximating $\sqrt{3}\approx 1.732$, $\tau \approx 0.005\times 1.732=0.00866$ N m. This is not an integer. Let's recheck the angle interpretation. Usually, 'angle with the direction of the magnetic field' refers to the angle of the normal. If it refers to the plane, then ${\theta }^{\prime }$ is $90-30=60$. If it means the angle between M and B is 30 degrees, then $\sin (30^{\circ })=0.5$. Let's assume the angle between the magnetic moment and the magnetic field is $30^{\circ }$.

If the angle between the plane of the loop and the magnetic field is $30^{\circ }$, then the angle between the magnetic moment vector (normal to the plane) and the magnetic field is $\theta =90^{\circ }-30^{\circ }=60^{\circ }$.$\tau =MB\sin \theta =(0.05{\text{ A m}}^2)\times (0.2\text{ T})\times \sin (60^{\circ })=0.01\times \frac{\sqrt{3}}{2}=0.005\sqrt{3}$ N m. This is not an integer in $10^{-3}$ N m.

Let's assume the question implies the angle between the magnetic moment and the magnetic field is $30^{\circ }$. This is a common ambiguity. In JEE, if 'plane makes angle $\alpha$ with B', then $\theta =90-\alpha$. If 'dipole makes angle $\alpha$ with B', then $\theta =\alpha$. Let's assume the angle between the magnetic moment and the magnetic field is $30^{\circ }$ for an integer answer.$\tau =MB\sin (30^{\circ })=(0.05{\text{ A m}}^2)\times (0.2\text{ T})\times 0.5=0.01\times 0.5=0.005$ N m. In $10^{-3}$ N m, the torque is $0.005\times 1000=5$ N m. The value of $N$ is 5.