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Question

Physics Question on Amperes circuital law

20 amp current is flowing in a long straight wire. The intensity of magnetic field at a distance of 10 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

Here:  i=20~i=20 amp, r=10 cm=10×10−2mr=10\text{ }cm=10\times {{10}^{-2}}m Intensity of magnetic field produced due to straight current carrying wire will be B=μ04π×2irB=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2i}{r} =10−7×2×200.1=4×10−5 wb/m2=\frac{{{10}^{-7}}\times 2\times 20}{0.1}=4\times {{10}^{-5}}\,wb/{{m}^{2}}