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MCQEasyJEE 2023Magnetic Dipole & Bar Magnet

JEE Physics 2023 Question with Solution

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JEE Main 2023 Questions with Solutions

A bar magnet with a magnetic moment 5.0 A m25.0 \, \text{A m}^2 is placed in parallel position relative to a magnetic field of 0.4 T0.4 \, \text{T}. The amount of required work done in turning the magnet from parallel to antiparallel position relative to the field direction is:

  • A

    4 J4 \, \text{J}

  • B

    1 J1 \, \text{J}

  • C

    2 J2 \, \text{J}

  • D

    Zero

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Magnetic moment M=5.0 A m2M = 5.0 \, \text{A m}^2 and magnetic field B=0.4 TB = 0.4 \, \text{T}.

Find: The work required to rotate the bar magnet from parallel to antiparallel position.

The potential energy of a magnetic dipole in a magnetic field is

u=−MBcos⁡θu = -MB\cos\theta

The work done in rotating the dipole is the change in potential energy:

W=ΔuW = \Delta u

Initially, the magnet is parallel to the field, so

θ=0∘\theta = 0^\circ

Finally, it is antiparallel to the field, so

θ=180∘\theta = 180^\circ

Therefore,

W=−MBcos⁡180∘−(−MBcos⁡0∘)W = -MB\cos 180^\circ - \left(-MB\cos 0^\circ\right) W=−MB(−1)−(−MB)(1)W = -MB(-1) - (-MB)(1) W=MB+MB=2MBW = MB + MB = 2MB

Substituting the values,

W=2×5×0.4W = 2 \times 5 \times 0.4 W=4 JW = 4 \, \text{J}

Therefore, the correct option is A.

Energy Change Approach

Given: A magnetic dipole is rotated in a uniform magnetic field from parallel to antiparallel orientation.

Find: The required work done.

Use the magnetic potential energy relation:

u=−MBcos⁡θu = -MB\cos\theta

For the initial parallel position,

ui=−MBcos⁡0∘=−MBu_i = -MB\cos 0^\circ = -MB

For the final antiparallel position,

uf=−MBcos⁡180∘=+MBu_f = -MB\cos 180^\circ = +MB

Hence the change in energy is

Δu=νf−νi\Delta u = \nu_f - \nu_i Δu=MB−(−MB)\Delta u = MB - (-MB) Δu=2MB\Delta u = 2MB

Since the external work done equals the increase in potential energy,

W=2MB=2×5.0×0.4=4 JW = 2MB = 2 \times 5.0 \times 0.4 = 4 \, \text{J}

Therefore, the answer is 4 J4 \, \text{J}.

Common mistakes

  • Using W=MBW = MB instead of W=2MBW = 2MB. This is wrong because the magnet moves from 0∘0^\circ to 180∘180^\circ, so the potential energy changes from −MB-MB to +MB+MB. Always compute final minus initial energy.

  • Confusing torque with work done. Torque depends on sin⁡θ\sin\theta, but the required work here must be found from the change in potential energy. Use u=−MBcos⁡θu = -MB\cos\theta.

  • Taking the final energy at 180∘180^\circ as negative. This is incorrect because cos⁡180∘=−1\cos 180^\circ = -1, hence u=−MB(−1)=+MBu = -MB(-1) = +MB. Evaluate the cosine carefully before substituting.

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