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A current of 3A is flowing in a linear conductor having a length of 40 cm. The conductor is placed in a magnetic field of strength 500 Gauss and makes an angle of $30 °$ with the direction of the field. It experiences a force of magnitude

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3 \times 10^{2} N

3 \times 10^{4} N

3 \times 10^{- 4} N

3 \times 10^{- 2} N

answer is C.

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Detailed Solution

i = 3 A, l = 40 cm, B = 500 G

$B = 500 \times 10^{- 4} T θ = 30°$

$F = B I l sin θ$

$= 500 \times 10^{- 4} \times 3 \times 40 \times 10^{- 2} \sin 30 °$

$= 3 \times 10^{- 2} N$

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