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IV. Numerical Problems · Q1

Q.A square coil of side 30 cm with 500 turns is kept in a uniform magnetic field of 0.4 T. The plane of the coil is inclined at an angle of 30∘30^{\circ} to the field. Calculate the magnetic flux through the coil.

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Step 1. Given: N = 500 turns, side = 30 cm = 0.3 m so A=0.32=0.09 m2A=0.3^2=0.09\ \text{m}^2, B=0.4B=0.4 T, plane inclined 30∘30^{\circ} to the field.

Step 2. The angle used in ΦB=BAcos⁡θ\Phi_B=BA\cos\theta is between the field and the NORMAL, so θ=90∘−30∘=60∘\theta=90^{\circ}-30^{\circ}=60^{\circ}.

Step 3. For N turns, total flux linkage is ΦB=NBAcos⁡θ=500×0.4×0.09×cos⁡60∘\Phi_B=NBA\cos\theta = 500\times0.4\times0.09\times\cos60^{\circ}.

Step 4. Computing: 500×0.4=200500\times0.4=200; 200×0.09=18200\times0.09=18; 18×0.5=918\times0.5=9.

✓Final answer

The magnetic flux through the coil is ΦB=9\Phi_B = 9 Wb.

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