The 10-turn loop of wire shown in FIGURE P29.71 lies in a horizontal plane, parallel to a uniform horizontal magnetic field, and carries a 2.0 A current. The loop is free to rotate about a nonmagnetic axle through the center. A 50 g mass hangs from one edge of the loop. What magnetic field strength will prevent the loop from rotating about the axle?

The diagram is as follows:

The loop of wire has $10$turns, the mass of the block is $50\text{g}$and the loop carries a current of$2.0\text{A}$.

If the loop is not allowed to rotate about the axle, the net torque created on the loop due to both the magnetic field and the block has to be equal with each other.

The formula to calculate the torque about the axle on the loop due to the block is given by

${\tau }_1=mgl...........................\left(1\right)$

Here, ${\tau }_1$is the torque due to the block, $m$is the mass of the block, $g$is the acceleration due to gravity and $l$is the distance of the block from the axle.

There is no torque on the arms of the loop which are parallel to the magnetic field. Only the arms which are at right angles to the direction of the magnetic field will experience the torque due to the magnetic field.

The formula to calculate the torque on the loop due to the magnetic field is given by

${\tau }_2=2nidBl........................\left(2\right)$

Here, ${\tau }_2$is the torque due to the magnetic field, $n$is the number of turns in the loop, $i$is the current through the loop,$d$is the length of each arm perpendicular to the magnetic field and$B$is the magnetic field.

Equate equation (1) and equation (2) and simplify to obtain the magnetic field.

$\begin{array}{rcl}mgl& =& 2nidBl\\ B& =& \frac{mg}{2nid}................................\left(3\right)\end{array}$

Substitute $50\text{g}$for $m$, $9.80{\text{m/s}}^2$for $g$, $10$for $n$,$2.0\text{A}$for $i$and $10.0\text{cm}$for $d$into equation (3) to calculate the required magnetic field.

$\begin{array}{rcl}B& =& \frac{50\text{g}\times {\displaystyle \frac{1\text{kg}}{1000\text{g}}}\times 9.80{\text{m/s}}^2}{2\times 10\times 2.0\text{A}\times 10.0\text{cm}\times {\displaystyle \frac{1\text{m}}{100\text{cm}}}}\\ & \approx & 0.12\text{T}\end{array}$

The required magnetic field is$\begin{array}{rcl}& & 0.12\text{T}\end{array}$.