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A horizontal overhead power line carries a current of 100A in east to west direction. What is the magnitude and direction of the magnetic field due to the current, 2 m below the line?(a) 3 x 10^-5 T(b) 2 x 10^-5 T(c) 1 x 10^-5 T(d) 4 x 10^-5 TThis question was posed to me during a job interview.This intriguing question comes from Magnetic Field on the Axis of a Circular Current Loop in section Moving Charges and Magnetism of Physics – Class 12

Answer»

Correct answer is (c) 1 x 10^-5 T

For explanation I would say: Given: I = 100A; r = 2 m; \(\frac {\mu_o}{4 \pi }\) = 10^-7

The required equation → B = \((\frac {\mu_o}{4 \pi }) \, \times \, \frac {2I}{r}\)

B = 10^-7 × 2 × \(\frac {100}{2}\)

B = 10^-7 × 100

B = 1 × 10^-5 T

Therefore, the magnitude of the MAGNETIC FIELD is 1 × 10^-5 T and the direction of the magnetic field is south.


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