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PhysicsMagnetismMagnetic torque & energyEasy2 minPYQ_2020

PhysicsEasynumerical

Calculate the couple (in $μ N m$ ) acting on a magnet of length $10 cm$ and pole strength $15 A m$ , kept in a field of $B = 2 \times 10^{- 5} T$ at an angle of $30 °$ .

Answer:

15.00

Solution:

$C = M B \sin θ$

$= (m \times 2 l) \times 2 \times 10^{- 5} \sin 30 °$

$= 15 \times 10 \times 10^{- 2} \times 2 \times 10^{- 5} \times \frac{1}{2}$

$= 15 \times 10^{- 6} N m$

Stream:JEESubject:PhysicsTopic:MagnetismSubtopic:Magnetic torque & energy

⏱ 2mℹ️ Source: PYQ_2020

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