Question

A circular coil with a radius of \(10.0 \mathrm{~cm}\) has 100 turns of wire and carries a current, \(i=100 . \mathrm{mA}\). It is free to rotate in a region with a constant horizontal magnetic field given by \(\vec{B}=(0.0100 \mathrm{~T}) \hat{x}\). If the unit normal vector to the plane of the coil makes an angle of \(30.0^{\circ}\) with the horizontal, what is the magnitude of the net torque acting on the coil?

Short answer

Question: Determine the net torque acting on a circular coil with a radius of 10.0 cm, 100 turns, a current of 100 mA, and a magnetic field of 0.0100 T. The angle between the unit normal vector and the horizontal is 30.0 degrees. Answer: The net torque acting on the circular coil is 100π N·cm.

Step-by-step solution

Calculating the area of the coil

To calculate the area (A) of the circular coil, we use the formula for the area of a circle: \(A = \pi r^2\), where r is the radius of the coil. In this case, the radius is given as \(10.0 \mathrm{~cm}\), so we have: \(A = \pi (10.0 \mathrm{~cm})^2 = 100\pi\, \mathrm{cm^2}\).

Converting the angle from degrees to radians

The angle between the unit normal vector and the horizontal, \(\theta\), is given in degrees (\(30.0^\circ\)). In order to use this angle in our calculation for the torque, we need to convert it to radians. To convert degrees to radians, we use the following formula: \(\mathrm{Radians} = \frac{\mathrm{Degrees}}{180^\circ} \cdot \pi\). Applying this to our given angle, we get: \(\theta = \frac{30.0^\circ}{180^\circ} \cdot \pi = \frac{\pi}{6}\, \mathrm{rad}\).

Calculating the net torque acting on the coil

Now, we will use the formula for torque on a current loop, which is \(\tau = N \cdot I \cdot A \cdot B \cdot \sin{\theta}\). We are given the number of turns (N = 100), the current (I = \(100 \mathrm{~mA}\) = \(0.1 \mathrm{~A}\)), the magnetic field (B = \(0.0100 \mathrm{~T}\)), and we have calculated the area (A = \(100\pi\, \mathrm{cm^2}\)) and the angle in radians (\(\theta = \frac{\pi}{6}\)). Plugging these values into the formula, we get: \(\tau = 100 \cdot 0.1 \mathrm{A} \cdot 100\pi\, \mathrm{cm^2} \cdot 0.0100 \mathrm{~T} \cdot \sin{\frac{\pi}{6}} = 100\pi\, \mathrm{N \cdot cm}\).