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Physics Question on Magnetic Field Due To A Current Element, Biot-Savart Law

Two concentric circular coils of ten turns each are situated in the same plane. Their radii are 20 cm20\, cm and 40 cm40\, cm and they carry respectively 0.2 A0.2\, A and 0.3 A0.3\, A currents in opposite direction. The magnetic field in tesla at the centre is

A

35μ04\frac{35\mu_{0}}{4}

B

μ080\frac{\mu_{0}}{80}

C

7μ080\frac{7 \mu_{0}}{80}

D

5μ04\frac{5 \mu_{0}}{4}

Answer

5μ04\frac{5 \mu_{0}}{4}

Explanation

Solution

Magnetic field at the centre of circular coil consists of N turns and radius r carrying current i is given by B=μ02πNi4πrB=\frac{\mu_{0}2\pi Ni}{4\pi r} For first coil, B1=μ02πN1i14πr1B_{1}=\frac{\mu_{0} 2\pi N_{1}i_{1}}{4\pi r_{1}} =μ0×10×0.22×0.2=\frac{\mu_{0}\times10\times0.2}{2\times0.2} For second coil, B2=μ02πN2i24πr2=μ0×10×0.32×0.4B_{2}=\frac{\mu_{0} 2\pi N_{2}i_{2}}{4\pi r_{2}}=\frac{\mu_{0}\times10\times0.3}{2\times0.4} The resultant magnetic field at the centre of the concentric coil is B=B1−B2=μ0×10×0.22×0.2−μ0×10×0.32×0.4B=B_{1}-B_{2}=\frac{\mu_{0}\times10\times0.2}{2\times0.2}-\frac{\mu_{0}\times10\times0.3}{2\times0.4} =10μ02[1−34]=10μ02(14)=5μ04=\frac{10\mu_{0}}{2} \left[1-\frac{3}{4}\right]=\frac{10\mu_{0}}{2}\left(\frac{1}{4}\right)=\frac{5\mu_{0}}{4}