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A circular loop of radius 11.7 cm is placed in a uniform magnetic field. (a) If the field is directed perpendicular to the plane of the loop and the magnetic flux through the loop is 8.60 10-3 T · m2, what is the strength of the magnetic field?

Respuesta :

Answer:

Magnetic field, B = 0.199 T

Explanation:

It is given that,

Radius of circular loop, r = 11.7 cm = 0.117 m

Magnetic flux through the loop, [tex]\phi=8.6\times 10^{-3}\ T/m^2[/tex]

The magnetic flux linked through the loop is :

[tex]\phi=B.A[/tex]

[tex]\phi=BA\ cos\theta[/tex]

Here, [tex]\theta=0[/tex]

[tex]B=\dfrac{\phi}{A}[/tex]

or

[tex]B=\dfrac{\phi}{\pi r^2}[/tex]

[tex]B=\dfrac{8.6\times 10^{-3}}{\pi (0.117 )^2}[/tex]

B = 0.199 T

So, the strength of the magnetic field is 0.199 T. Hence, this is the required solution.

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