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in Electromagnetic induction and alternating currents by (25 points)
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A uniform magnetic field exists along \( x \)-axis, find the flux through a square loop of side a inclined at \( 60^{\circ} \) to \( (+x) \) axis as shown in figure. 

(1) \( \frac{B a^{2} \sqrt{3}}{2} \) 

(2) \( \frac{B a^{2}}{2} \)

(3) \( B a^{2} \) 

(4) Zero

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1 Answer

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by (53.5k points)
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Correct option is (1) \(\frac{Ba^2\sqrt 3}2\)

We know that 

flux \(\phi = B.A\)

Now, 

Area = a2 in 30° with x-axis

Because area is always perpendicular to plane of loop

\(\phi= B.A \,cos60°\)

\(Ba^2 \frac{\sqrt3}2\)

\(\phi = \frac{\sqrt 3}2 Ba^2\)

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