A magnet of length 0.1 m and pole strength $10^{- 4} A . m$ is kept in a magnetic field of $30 W b / m^{2}$ at an angle $30^{\circ}$ . The couple acting on it is ……… $\times 10^{- 4} N m$ .

7.5

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1.5

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Solution

The correct option is C 1.5
$τ = M B s i n θ = m \times (2 l) \times B s i n θ$
$= 10^{- 4} \times 0.1 \times 30 s i n 30^{\circ} = 1.5 \times 10^{- 4} N m$

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A magnet of length 0.1 m and pole strength 10−4 A.m is kept in a magnetic field of 30 Wb/m2 at an angle 30∘. The couple acting on it is ……… ×10−4 Nm.

The correct option is C 1.5τ=MB sin θ=m×(2l)×B sin θ =10−4×0.1×30 sin 30∘=1.5×10−4 Nm

A magnet of length 0.1 m and pole strength $10^{- 4} A . m$ is kept in a magnetic field of $30 W b / m^{2}$ at an angle $30^{\circ}$ . The couple acting on it is ……… $\times 10^{- 4} N m$ .

The correct option is C 1.5
$τ = M B s i n θ = m \times (2 l) \times B s i n θ$
$= 10^{- 4} \times 0.1 \times 30 s i n 30^{\circ} = 1.5 \times 10^{- 4} N m$