91Ó°ÊÓ

1. The number of turns of the coil is, N = 160 .
2. The radius of the coil is, r = 1.90 cm = 0.019 m .
3. The magnetic dipole moment of the coil is, $\mathrm{μ}=2.30A.m^2$.
4. The external magnetic field is, $B=35.0mT=35.0×{10}^{-3}T$.

When a coil carries current, it creates a magnetic dipole moment. We can find the current through the coil by substituting the given values in the formula for magnetic dipole moment. The torque acting on the coil due to external magnetic field is given by the cross product of the magnetic dipole moment of the loop and the external magnetic field.

$$\begin{gathered} \mathit{μ}\mathbf{=}\mathit{N}\mathit{i}\mathit{A} \\ \mathit{Ï„}\mathbf{=}\stackrel{\mathbf{â‡\P Ä}}{\mathbf{μ}}\mathbf{×}\stackrel{\mathbf{â‡\P Ä}}{\mathbf{B}}\mathbf{=}\mathit{μ}\mathit{s}\mathit{i}\mathit{n}\mathit{θ} \end{gathered}$$

The magnetic moment is calculated as

$\mathrm{μ}=NiA$

For, $A=\mathrm{Ï€}{r}^2$,

Then,

$\mathrm{μ}=Ni{\mathrm{π}}^2$

Rearranging the equation,

$i=\frac{\mathrm{μ}}{N\mathrm{π}{r}^2}$

Substitute all the value in the above equation.

$\begin{array}{rcl}i& =& \frac{2.30A{m}^2}{160×3.14×{\left(0.019m\right)}^2}\\ & =& 12.7A\end{array}$

Hence the value of current is, $\begin{array}{rcl}& & 12.7A\end{array}$.

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

The torque is calculated as

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

The magnitude of the torque will be maximum when $\mathrm{θ}=1$, i.e. when $\mathrm{θ}={90}^o$

$\mathrm{τ}=\mathrm{μ}B$

Substitute all the value in the above equation.

$$\begin{gathered} \mathrm{τ}=2.30A{m}^2×35.0×{10}^{-3}T \\ \mathrm{τ}=0.0805Nm \end{gathered}$$

Hence the maximum magnitude of the torque is, 0.0805Nm .