A circular loop of radius $R=20 cm$ is placed in a uniform magnetic field $B=2 T$ in x-y plane as shown in the figure. The loop carries a current $i=1.0 A$ in the direction shown in the figure. Fidn the magnitude of torque acting on the loop.

Magnitude of torque is given by $\left|\tau \right|=MB\sin \theta$

Here, $M=NiA$

             $=\left(1\right)\left(1.0\right)\left(\mathrm{\pi }\right){\left(0.2\right)}^2$

             $=\left(0.04\mathrm{\pi }\right) A-m^2$

$B=2 T$ and $\theta$= angle between $M$ and $B=90°$

$$\begin{gathered} \left|\tau \right|=\left(0.04\mathrm{\pi }\right)\left(2\right)\sin 90° \\ =0.25 N-m \end{gathered}$$

$M$ is along negative $z$-direction (perpendicular to paper inwards) while $B$ is in $x-y$ plane. So, the angle between $M$ and $B$ is $90°$ not $45°$. If the direction of torque is also desired, then we can write

$$\begin{gathered} B=2\cos 45°\hat{i}+2\sin 45°\hat{j} \\ =\sqrt{2}\left(\hat{i}+\hat{j}\right)T \\ M=-\left(0.04\mathrm{\pi }\right)\hat{k} A-m^2 \\ \tau =M\times B \\ =\left(0.04\sqrt{2}\mathrm{\pi }\right)\left(-\hat{j}+\hat{i}\right) \\ =0.18\left(\hat{i}-\hat{j}\right) \end{gathered}$$