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Exercise · Q10

Q.A plane loop of area AA is placed in a uniform magnetic field BB such that the normal to the loop makes an angle θ\theta with BB. Write the general expression for the flux linked with the loop, and use it to state the flux when

(a) the plane of the loop is perpendicular to BB and
(b) the plane of the loop is parallel to BB.
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✓ Free question

The flux linked with a plane loop of area AA, whose normal makes angle θ\theta with a uniform field BB, is

ΦB=BAcos⁡θ\Phi_B = BA\cos\theta

(a) Plane of the loop perpendicular to BB. Here the loop's normal is PARALLEL to BB, so the angle between the normal and BB is θ=0∘\theta=0^\circ, giving cos⁡θ=1\cos\theta=1 and

ΦB=BA(maximum possible flux)\Phi_B = BA \quad \text{(maximum possible flux)}

(b) Plane of the loop parallel to BB. Here the loop's normal is PERPENDICULAR to BB, so θ=90∘\theta=90^\circ, giving cos⁡θ=0\cos\theta=0 and

ΦB=0\Phi_B = 0

no field lines cross the loop's surface at all in this orientation.

✓Final answer

Maximum flux ΦB=BA\Phi_B=BA when the loop's plane is perpendicular to BB; zero flux when the loop's plane is parallel to BB.

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