A circular loop of radius$\mathbf{\text{12cm}}$carries a current of$\mathbf{\text{15A}}$. A flat coil of radius$\mathbf{\text{0.82cm}}$having$\mathbf{50}\mathbf{\text{turns}}$and a current of $\mathbf{1}\mathbf{.3}\mathbf{\text{A}}$, is concentric with the loop. The plane of the loop is perpendicular to the plane of the coil. Assume the loop’s magnetic field is uniform across the coil. (a) What is the magnitude of the magnetic field produced by the loop at its center and (b) What is the magnitude of the torque on the coil due to the loop?

• The loop radius,$r=12\text{\hspace{0.17em}cm}$
• The current, $i=15\text{\hspace{0.17em}A}$
• The flat coil radius, $R=0.82\text{cm}$
• The number of turns, $N=50$
• The current through coil, $I=1.3\text{\hspace{0.17em}A}$

Define the magnetic field due to the circular loop at center by using the formula in terms of current and radius. Then, you use this magnetic field to find torque using the relation between torque, magnetic field, and magnetic dipole moment.

Formulae:

The magnetic field is defined by,

$B=\frac{{\mu }_{0}i}{2r}$

The magnetic dipole moment is,

$\mu =NiA$

Here, ${\mu }_0$is the permeability of free space having a value $4\pi \times {10}^{-7}\raisebox{1ex}{${\displaystyle \text{Tm}}$}\!\left/ \!\raisebox{-1ex}{$\text{A}$}\right.$,$i$is the current,$I$is the current through coil,$r$is the radius, $B$is the magnetic field, $N$is the number of turns, and $A$is the area.

Define the magnetic field at center as follows.

$B=\frac{\mu i}{2r}$

Substitute all the value in the above equation.

$\begin{array}{rcl}B& =& \frac{4\pi \times {10}^{-7}\text{Tm}/\text{A}\times 15\text{A}}{2\times 0.12\text{m}}\\ & =& 7.9\times {10}^{-5}\text{T}\end{array}$

Hence, the magnetic field is role="math" localid="1663248173578" $7.9\times {10}^{-5}\text{T}$.

Now torque acting on coil of current $I$due to the loop with magnetic field $B$is as follows.

$\begin{array}{rcl}\tau & =& NIAB\sin \theta \\ & =& N\times I\times \pi \times R^2\times B\times \sin 90°\end{array}$

Substitute all the value in the above equation.

$\begin{array}{rcl}\tau & =& 50\times 1.3\text{A}\times 3.14\times {\left(0.0082\text{m}\right)}^2\times (7.9\times {10}^{-5}\text{T})\times 1\\ & =& 1.1\times {10}^{-6}\text{N}\cdot \text{m}\end{array}$

Hence, the torque is $1.1\times {10}^{-6}\text{N}\cdot \text{m}$.