The magnetic flux linked with a coil of N tur | vedclass

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Question

(D) The magnetic flux $\phi$ linked with a coil is given by the formula $\phi = NBA \cos 	heta$,where $	heta$ is the angle between the magnetic fiel

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zero

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(D) The magnetic flux $\phi$ linked with a coil is given by the formula $\phi = NBA \cos 	heta$,where $	heta$ is the angle between the magnetic field vector $\vec{B}$ and the area vector $\vec{A}$ (which is normal to the plane of the coil).Given that the plane of the coil is parallel to the magnetic field $B$,the angle between the normal to the coil and the magnetic field is $	heta = 90^{\circ}$.Substituting this into the formula: $\phi = NBA \cos(90^{\circ})$.Since $\cos(90^{\circ}) = 0$,the magnetic flux $\phi = 0$.

The magnetic flux linked with a coil of $N$ turns of area of cross section $A$ held with its plane parallel to the field $B$ is

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