Question

A circular coil of wire with 20 turns and a radius of \(40.0 \mathrm{~cm}\) is laying flat on a horizontal tabletop as shown in the figure. There is a uniform magnetic field extending over the entire table with a magnitude of \(5.00 \mathrm{~T}\) and directed to the north and downward, making an angle of \(25.8^{\circ}\) with the horizontal. What is the magnitude of the magnetic flux through the coil?

Short answer

Answer: The magnitude of the magnetic flux through the coil is approximately 2.40 Wb.

Step-by-step solution

Find the area of the coil

The coil has a radius of 40 cm or 0.4 m. The formula for the area of a circle is given by: \(A = \pi r^2\) We will substitute the radius into the formula to find the area enclosed by the coil: \(A = \pi (0.4\,\text{m})^2 = 0.16\pi\,\text{m}^2\)

Convert the angle to radians

We are given the angle at which the magnetic field is with the horizontal as \(25.8^{\circ}\). We need to convert this angle into radians: \(\theta = 25.8 \times \frac{\pi}{180} = 0.45\,\text{radians}\)

Calculate the angle between the magnetic field and the normal vector of the coil's plane

Since the coil is laying flat on the table, the normal vector of the coil's plane is upward and perpendicular to the tabletop (90 degrees from the horizontal). To find the angle between the magnetic field and the normal vector of the coil's plane, we need to subtract \(\theta\) from 90 degrees: \(\alpha = 90^{\circ} - 25.8^{\circ} = 64.2^{\circ}\) Now convert this angle into radians: \(\alpha = 64.2 \times \frac{\pi}{180} = 1.12\,\text{radians}\)

Calculate the magnetic flux through the coil

Now that we have the angle \(\alpha\) and the area \(A\), we can use the formula for magnetic flux to find the magnitude of magnetic flux through the coil: \(\Phi_B = BA\cos{\alpha}\) \(\Phi_B = (5.0\,\text{T})(0.16\pi\,\text{m}^2) \cos(1.12\,\text{radians})\) \(\Phi_B \approx 2.40\,\text{Wb}\) (rounded to two decimal places) The magnitude of the magnetic flux through the coil is approximately 2.40 Wb.