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PhysicsMagnetismMagnetic Dipole & MomentHard2 minPYQ_2024

PhysicsHardnumerical

A coil having 100 turns, area of $5 \times 1 0^{- 3} m^{2}$ , carrying current of $1 mA$ is placed in uniform magnetic field of $0.20 T$ such a way that plane of coil is perpendicular to the magnetic field. The work done in turning the coil through $9 0^{\circ}$ is _______ $μ J$ .

Answer:

100.00

Solution:

$W = Δ U = U_{f} - U_{i} W = (- μ . B)_{f} - (- μ . B)_{i} = 0 + (μ \cdot B)_{i} = (100 \times 5 \times 1 0^{- 3} \times 1 \times 1 0^{- 3}) \times 0.2 J = 1 \times 1 0^{- 4} J = 100 μ J$

Stream:JEESubject:PhysicsTopic:MagnetismSubtopic:Magnetic Dipole & Moment

⏱ 2mℹ️ Source: PYQ_2024

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