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Question

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

20 A20 \,A current is flowing in a long straight wire. The intensity of magnetic field at a distance of 10 cm10\, cm from the wire, will be

A

4×10−5Wb/m24 \times 10^{-5} Wb / m ^{2}

B

2×10−5Wb/m22 \times 10^{-5} Wb / m ^{2}

C

3×10−5Wb/m23 \times 10^{-5} Wb / m ^{2}

D

8×10−5Wb/m28 \times 10^{-5} Wb / m ^{2}

Answer

4×10−5Wb/m24 \times 10^{-5} Wb / m ^{2}

Explanation

Solution

Intensity of magnetic field produced due to straight current carrying wire will be
B=μ04π2IrB=\frac{\mu_{0}}{4 \pi} \frac{2 I}{r}
Given, I=20 AI=20\, A,
r=10 cmr =10\, cm
=10×10−2m=10 \times 10^{-2} m
∴B=10−7×2×2010×10−2\therefore B =10^{-7} \times \frac{2 \times 20}{10 \times 10^{-2}}
=4×10−5Wb/m2=4 \times 10^{-5} Wb / m ^{2}