91Ó°ÊÓ

The radius of the wire loop = $r=15.0cm=0.15m$

The current through the loop = $i=2.60A$

The angle between the normal to the plane of the loop and the external magnetic field = $\mathrm{θ}=41.0^o$

The external magnetic field =$B=12.0T$

The magnetic dipole moment of a current carrying loop depends on the number of loops, the current through the loop, and the area enclosed by each turn of the loop. Substituting these values in the formula, we can find the magnitude of the magnetic dipole moment. The torque acting on the loop due to external magnetic field is given by the cross product of the magnetic moment of the loop and the external magnetic field.

$$\begin{gathered} \mathrm{μ}=NiA \\ \mathrm{Ï„}=\stackrel{â‡\P Ä}{\mathrm{μ}}×\stackrel{â‡\P Ä}{B}=\mathrm{μ}B\sin \mathrm{θ} \end{gathered}$$

The magnetic moment is given by

$\mathrm{μ}=NiA$

Here,$N=1andA=\mathrm{Ï€}{r}^2$

$$\begin{gathered} \mathrm{μ}=i\mathrm{π}{r}^2 \\ \mathrm{μ}=2.60×3.14×\left(0.{15}^2\right) \\ \mathrm{μ}=2.60×3.14×\left(0.0225\right) \\ \mathrm{μ}=0.184A.m^2 \end{gathered}$$

Hence,$\mathrm{μ}=0.184A.m^2$

When the current carrying loop is placed in an external magnetic field, it experiences a torque.

The magnitude of the torque is calculated as

$$\begin{gathered} \stackrel{â‡\P Ä}{\mathrm{Ï„}}=\stackrel{â‡\P Ä}{\mathrm{μ}}×\stackrel{â‡\P Ä}{B} \\ \mathrm{Ï„}=\mathrm{μ}B\sin \mathrm{θ} \\ \mathrm{Ï„}=0.184×12.0×\sin \left(41.0\right) \\ \mathrm{Ï„}=0.184×12.0×0.6561 \\ \mathrm{Ï„}=1.45N·m \end{gathered}$$

Hence,$\mathrm{τ}=1.45N·m$