Question

A rectangular coil with 20 windings carries a current of \(2.00 \mathrm{~mA}\) flowing in the counterclockwise direction. It has two sides that are parallel to the \(y\) -axis and have length \(8.00 \mathrm{~cm}\) and two sides that are parallel to the \(x\) -axis and have length \(6.00 \mathrm{~cm}\). A uniform magnetic field of \(50.0 \mu \mathrm{T}\) acts in the positive \(x\) -direction. What torque must be applied to the loop to hold it steady?

Short answer

Answer: The external torque required to hold the loop steady is -4.80 x 10^-6 Nm.

Step-by-step solution

Magnetic force on each side of the loop

To find the magnetic force on each side of the loop, we use the formula \(F = I L B \sin(\theta)\), where \(F\) is the force, \(I\) is the current, \(L\) is the length of the side, \(B\) is the magnetic field, and \(\theta\) is the angle between the current's direction and the magnetic field's direction. For the sides parallel to the y-axis, the current is flowing perpendicular to the magnetic field, so \(\theta = 90^{\circ}\). Thus, the sine of the angle is 1. For the sides parallel to the x-axis, the current is flowing parallel to the magnetic field, so \(\theta = 0^{\circ}\). Thus, the sine of the angle is 0.

Calculate the force on each side

Next, we calculate the force on each side using the formula \(F = I L B \sin(\theta)\). Sides parallel to the y-axis: \(F_y = (2.00 \times 10^{-3} \mathrm{A})(8.00 \times 10^{-2} \mathrm{m})(50.0 \times 10^{-6} \mathrm{T})(1) = 8.00 \times 10^{-6} \mathrm{N}\) Sides parallel to the x-axis: \(F_x = 0 \text{ N}\). Since sine of \(\theta\) is 0, there is no magnetic force acting on these sides.

Calculate the net torque

To find the net torque about the center of the loop, we first determine the distance from the center of the loop to the y-axis sides. The distance is half the length of the x-axis sides, \(d = (6.00 \times 10^{-2} \mathrm{m}) / 2 = 3.00 \times 10^{-2} \mathrm{m}\). The torque \(\tau\) is calculated as \(\tau = F_y \cdot d\). Since the forces act in opposite directions on the top and bottom sides, they create a torque in the same direction. Hence, we multiply the torque from one side by the number of windings (20) to get the total torque. \(\tau = 20(8.00 \times 10^{-6} \mathrm{N})(3.00 \times 10^{-2} \mathrm{m}) = 4.80 \times 10^{-6} \mathrm{Nm}\)

External torque required

The external torque required to hold the loop steady will be equal in magnitude but opposite in direction to the torque calculated in the previous step. External torque required: \(-4.80 \times 10^{-6} \mathrm{Nm}\)