Question

A circular area with a radius of 6.50 cm lies in the xy-plane. What is the magnitude of the magnetic flux through this circle due to a uniform magnetic field B = 0.230 T (a) in the +z-direction; (b) at an angle of 53.1∘ from the +z-direction; (c) in the +y-direction?

Answer 1

Answer:

$a. \phi _B=3.0528\times10^-^3T\ m^2\\ \\b. \phi_B=1.83299\times10^-^3T\ m^2\\\\c.\phi_B=0$

Explanation:

#Consider a circular area of radius $R=2.98cm$ in the xy-plane at z=0. This means all the are vector points toward the +ve z-axis.

a. first, find the magnetic flux if the magnetic field has a magnitude of $B=0.23T$ and points toward the +ve z-axis. The angle between the magnetic field and the area is $\theta=0$. Hence the magnetic flux:-

$\phi _B=\int {\bar B} . d\bar A \\=\int BdAcos(\theta)=BAcos(0)=BA\\\\=\pi R^2B=\pi(6.50\times10^-^3m)^2(0.230T)\\=3.0528\times10^-^3 T\ m^2$

Hence flux magnitude in $+z$ direction is $3.0528\times10^-^3T \ m^2$

b. We now find the magnetic flux when the field has a magnitude of B=0.230T and points at an angle of $\theta=53.1\textdegree$ from the $+z$ direction.

Magnetic flux is calculated as:

$\phi _B=\int\bar B \bar dA\\=\int BdAcos (\theta)=BAcos(0)=BA\\=\pi R^2B=\pi(6.50\times 10^-^2m)^2(0.230T)\\=1.83299\times 10^-^3 T \ m^2$

Hence the flux at an angle of $53.1\textdegree$ is $1.83299\times 10^-^3T \ m^2$

c. We now need to find the magnetic flux if the field has a magnitue of B=0.230T and points in the direction of +y-direction. As with the previous parts, the magnetic flux will be calculated as:

$\phi_B= \int\bar B \times d\bar A\\=\int BdAcos(\theta)\\=BAcos(90\textdegree)\\=0$

Hence the magnetic flux in the +y-direction is zero.