Question

A circular coil that has N = 180 N=180 turns and a radius of r = 13.0 cm r=13.0 cm lies in a magnetic field that has a magnitude of B 0 = 0.0790 T B0=0.0790 T directed perpendicular to the plane of the coil. What is the magnitude of the magnetic flux Φ B ΦB through the coil?

Answer 1

Answer:

The magnitude of the magnetic flux  through the coil is $0.75\ T-m^2$.

Explanation:

Number of turns in a circular coil, N = 180

Radius of the coil, r = 13 cm = 0.13 m

Magnitude of magnetic field, B = 0.079 T

The magnetic field is directed perpendicular to the plane of the coil. We need to find the magnitude of the magnetic flux  through the coil. The magnetic flux is given by the formula as :

$\phi=NBA\ \cos\theta$

$\theta=0$

$\phi=180\times 0.079\times \pi (0.13)^2\\\\\phi=0.75\ T-m^2$

So, magnitude of the magnetic flux  through the coil is $0.75\ T-m^2$. Hence, this is the required solution.