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Question 89 of 91

Q.The potential energy of a magnetic dipole whose dipole moment is p⃗m=(−0.5i^+0.4j^) Am2\vec{p}_m = (-0.5\hat{i} + 0.4\hat{j})\ \text{Am}^2 kept in uniform magnetic field B⃗=2j^ T\vec{B} = 2\hat{j}\ \text{T} :

(a) 0.1 J
(b) −0.1 J
(c) 0.8 J
(d) −0.8 J
Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2026MCQ· 1mImportance★★★★★
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Taking the dot product p⃗m⋅B⃗\vec p_m\cdot\vec B and negating it gives a potential energy of −0.8-0.8 J.

Working

Potential energy of a magnetic dipole in a field: U=−p⃗m⋅B⃗U = -\vec p_m\cdot\vec B.

Given p⃗m=(−0.5i^+0.4j^) Am2\vec p_m=(-0.5\hat i+0.4\hat j)\ \text{Am}^2, B⃗=2j^ T\vec B=2\hat j\ \text{T}: …

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