A long wire carries a current of 90A from east to west direction. Magnetic field due to current at a point which is 1.5 m just below the wire.

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Solution

A long wire carries a current
$I = 90 A$
(From east to west direction)
Magnetic field due to current at the point which is $= 1.5 m$
$B = \frac{μ_{0} I}{2 π r} = \frac{4 π \times 10^{- 7} \times 90}{2 π \times 1.5} = 1.2 \times 10^{- 5} J$

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A long wire carries a current of 90A from east to west direction. Magnetic field due to current at a point which is 1.5 m just below the wire.

A long wire carries a current I=90A(From east to west direction)Magnetic field due to current at the point which is =1.5mB=μ0I2πr=4π×10−7×902π×1.5=1.2×10−5J

A long wire carries a current
$I = 90 A$
(From east to west direction)
Magnetic field due to current at the point which is $= 1.5 m$
$B = \frac{μ_{0} I}{2 π r} = \frac{4 π \times 10^{- 7} \times 90}{2 π \times 1.5} = 1.2 \times 10^{- 5} J$