Question

A 45.5-turn circular coil of radius 4.95 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0.460 T. If the coil carries a current of 24.3 mA, find the magnitude of the maximum possible torque exerted on the coil.

Answer 1

Concepts and reason

The concept used to solve this problem is the torque on coil in uniform magnetic field.

First, write the expression of torque on coil in uniform magnetic field and substitute the known values in the expression to calculate maximum possible toque exerted on the coil.

Fundamentals

The expression of torque exerted on a coil is given by,

$\tau = NIAB\sin \theta$

Here, $N$ is number of turns, $A$ is area of coil, $I$is current flowing through the conductor, $B$ is magnetic field and $\theta$ is the angle between magnetic field and areal vector.

The expression of torque exerted on a coil is given by,

$\tau = NIAB\sin \theta$

Here, $N$ is number of turns, $A$ is area of coil, $I$is current flowing through the conductor, $B$ is magnetic field and $\theta$ is the angle between magnetic field and areal vector.

The maximum value of torque is possible at $\theta = 90^\circ$.

Substitute ${90^{\rm{o}}}$ for $\theta$ in equation $\tau = NIAB\sin \theta$.

$\tau = NIAB$

Now substitute $\pi {r^2}$ for $A$,

$\tau = NI\left( {\pi {r^2}} \right)B$

Substitute $45.5$for $N$, $4.95 {\rm{cm}}$ for $r$, $24.3 {\rm{mA}}$ for $I$ and $0.460 {\rm{T}}$ for $B$in equation $\tau = NI\left( {\pi {r^2}} \right)B$.

$\begin{array}{c}\\\tau = \left( {45.5 } \right)\left( {24.3 {\rm{mA}}} \right)\left( {\frac{{{{10}^{ - 3}} {\rm{A}}}}{{{\rm{1}} {\rm{mA}}}}} \right)\left( {\pi {{\left( {\left( {4.95 {\rm{cm}}} \right)\left( {\frac{{{{10}^{ - 2}} {\rm{m}}}}{{1 {\rm{cm}}}}} \right)} \right)}^2}} \right)\left( {0.460 {\rm{T}}} \right)\\\\ = 3.92 \times {10^{ - 3}} {\rm{N}} \cdot {\rm{m}}\\\end{array}$

Ans:

The possible maximum value of torque exerted on coil is $3.92 \times {10^{ - 3}} {\rm{N}} \cdot {\rm{m}}$.