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

Two concentric coils each of radius equal to 2 π2\, \pi cm are placed right angles to each other.If 3 A3\,A and 4 A4\,A are the currents flowing through the two coils respectively. The magnetic induction (in Wb m−2)(in\, Wb\, m^{-2}) at the centre of the coils will be

A

12×10−512\times 10^{-5}

B

10×10−510\times 10^{-5}

C

5×10−55\times 10^{-5}

D

7×10−57 \times 10^{-5}

Answer

5×10−55\times 10^{-5}

Explanation

Solution

Given, I1=3 AI_{1} =3\, A I2=4AI_{2} =4 A R=2πcm=2π×10−2 mR =2 \pi cm =2 \pi \times 10^{-2}\, m We know that, magnetic field of a coil, B=μ0I2RB=\frac{\mu_{0} I}{2 R} Now, B1=μ0I12R=μ02×32π×10−2B_{1} =\frac{\mu_{0} I_{1}}{2 R}=\frac{\mu_{0}}{2} \times \frac{3}{2 \pi \times 10^{-2}} =μ04π×310−2=\frac{\mu_{0}}{4 \pi} \times \frac{3}{10^{-2}} =10−7×310−2=3×10−5 T=10^{-7} \times \frac{3}{10^{-2}}=3 \times 10^{-5}\, T Similarly, B2=μ0I22R=μ02×42π×10−2B_{2} =\frac{\mu_{0} I_{2}}{2 R}=\frac{\mu_{0}}{2} \times \frac{4}{2 \pi \times 10^{-2}} =μ04π×410−2=\frac{\mu_{0}}{4 \pi} \times \frac{4}{10^{-2}} =10−7×4102=4×10−5 T=10^{-7} \times \frac{4}{10^{2}}=4 \times 10^{-5} \,T Now, net magnetic field at centre of a coil, B=B12+B22B=\sqrt{B_{1}^{2}+B_{2}^{2}} B=(3×10−5)2+(4×10−5)2B=\sqrt{\left(3 \times 10^{-5}\right)^{2}+\left(4 \times 10^{-5}\right)^{2}} B=10−10(9+16)B=\sqrt{10^{-10}(9+16)} B=5×10−5 TB=5 \times 10^{-5}\, T