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Standard XII

Physics

Torque

A magnet of l...

Question

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 $30^{o}$ . The couple acting on it is ……… $10^{- 4} N m$

7.5

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6.0

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Solution

The correct option is B $6.0$
Given
Pole strength $m = 10^{- 4} A m$ and length = $0.1 m$
$τ = \to M \times \to B = M B s i n θ = m \times (2 l) \times B s i n θ = 10^{- 4} \times 0.1 \times 30 s i n 30^{o} = 1.5 \times 10^{- 4} N m$ .

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

The correct option is B 6.0Given Pole strength m=10−4Am and length = 0.1mτ=→M×→B=MBsinθ=m×(2l)×Bsinθ=10−4×0.1×30sin30o=1.5×10−4Nm.

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 $30^{o}$ . The couple acting on it is ……… $10^{- 4} N m$

The correct option is B $6.0$
Given
Pole strength $m = 10^{- 4} A m$ and length = $0.1 m$
$τ = \to M \times \to B = M B s i n θ = m \times (2 l) \times B s i n θ = 10^{- 4} \times 0.1 \times 30 s i n 30^{o} = 1.5 \times 10^{- 4} N m$ .