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

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× 10-5T towards north (B)  3× 10-5T towards east (C)  1.2× 10-5T towards south (D)  1× 10-5T towards west

Tardigrade Admin · Accepted Answer

Magnetic field due to straight current carrying conductor is given by B=(μ 0/4π )2I/R =10-7× 2× 90/1.5=1.2× 10-5T towards south.

Q. 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

$2.5 \times 10^{- 5} T$ towards north

$3 \times 10^{- 5} T$ towards east

$1.2 \times 10^{- 5} T$ towards south

$1 \times 10^{- 5} T$ towards west

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

Q. 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

2.5×10−5T towards north

3×10−5T towards east

1.2×10−5T towards south

1×10−5T towards west

Magnetic field due to straight current carrying conductor is given by B=(μ0​/4π)2I/R =10−7×2×90/1.5=1.2×10−5T towards south.

$2.5 \times 10^{- 5} T$ towards north

$3 \times 10^{- 5} T$ towards east

$1.2 \times 10^{- 5} T$ towards south

$1 \times 10^{- 5} T$ towards west

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