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 unifo

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?

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Minh Khuê 5 months 2021-08-29T20:59:56+00:00 1 Answers 113 views 0

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    2021-08-29T21:01:48+00:00

    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.

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