Magnetic Effects & Magnetism

JEE Main level, single correct, medium

74 of 276

Question 74 of 276

Magnitude of magnetic field (in SI units) at the centre of a hexagonal shape coil of side 10cm,5010cm,50 turns and carrying currentII (Ampere) in units of μ0Iπ\dfrac{{\mu }_{0}I}{\pi } is :

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Answer: (C) 5003500\sqrt{3}

The magnetic field due to the current-carrying finite length wire is given by
B=μ0I4πR[sin⁡θ1+sin⁡θ2]B=\dfrac{{\mu }_{0}I}{4\pi R}\left[\sin {\theta }_{1}+\sin {\theta }_{2}\right]
Here for hexagonal shape coil, θ1=θ2=30∘{\theta }_{1}={\theta }_{2}=30^{\circ}.
So, the net magnetic field at the centre will be B=50×6×μ0i4π(10100cos⁡30∘)[sin⁡30∘+sin⁡30∘]B=50\times 6\times \dfrac{{\mu }_{0}i}{4\pi \left(\dfrac{10}{100}\cos 30^{\circ}\right)}\left[\sin 30^{\circ}+\sin 30^{\circ}\right]
B=2×75×10μ0i3π(12+12)B=2\times 75\times 10\dfrac{{\mu }_{0}i}{\sqrt{3}\pi }\left(\dfrac{1}{2}+\dfrac{1}{2}\right)
B=15003μ0iπ=(5003)μ0iπB=\dfrac{1500}{\sqrt{3}}\dfrac{{\mu }_{0}i}{\pi } =\left(500\sqrt{3}\right)\dfrac{{\mu }_{0}i}{\pi }

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