Question

A horizontal overhead power line carries a current of, 90 A in east to west direction. The magnitude and direction of the magnetic field due to the current 1.5m below the line is

• A. $ 2.5\times {{10}^{-5}}T $ towards north
• B. $ 3\times {{10}^{-5}}T $ towards east
• C. $ 1.2\times {{10}^{-5}}T $ towards south
• D. $ 1\times {{10}^{-5}}T $ towards west

Answer 1

Answer: (C)

Magnetic field due to straight current carrying conductor is given by $ B=({{\mu }_{0}}/4\pi )2I/R $ $ ={{10}^{-7}}\times 2\times 90/1.5=1.2\times {{10}^{-5}}T $ towards south.