A vertical rectangular coil of sides $5 c m \times 2 c m$ has $10$ turns and carries a current of $2 A$ . The torque(couple) on the coil when it is placed in a uniform horizontal magnetic field of $0.1 T$ with its plane perpendicular to the field is

4 \times 10^{- 3} N - m

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2 \times 10^{- 3} N - m

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10^{- 3} N - m

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Solution

The correct option is D $0$
If plane is perpendicular to the field, normal vector will be parallel to field
$sin θ = 0$
$J = M B sin θ = 0$

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A vertical rectangular coil of sides 5cm×2cm has 10 turns and carries a current of 2A. The torque(couple) on the coil when it is placed in a uniform horizontal magnetic field of 0.1T with its plane perpendicular to the field is

The correct option is D 0If plane is perpendicular to the field, normal vector will be parallel to field sinθ=0J=MBsinθ=0

A vertical rectangular coil of sides $5 c m \times 2 c m$ has $10$ turns and carries a current of $2 A$ . The torque(couple) on the coil when it is placed in a uniform horizontal magnetic field of $0.1 T$ with its plane perpendicular to the field is

The correct option is D $0$
If plane is perpendicular to the field, normal vector will be parallel to field
$sin θ = 0$
$J = M B sin θ = 0$