A planar coil having 10 turns and carrying a current of 10 A with an area vector \( \vec{A} = -0.04\,\hat{\imath}\;m^{2} \), is placed in a uniform magnetic field \( \vec{B} = 0.5\,\hat{\imath}\;T \). The potential energy of the coil is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2 J
A current-carrying coil behaves as a magnetic dipole with moment \( \vec{m} = NI\vec{A} \), and its potential energy in a uniform field is \( U = -\vec{m}\cdot\vec{B} \).
\( \displaystyle \vec{m} = NI\vec{A} = 10 \times 10\,\mathrm{A} \times (-0.04\,\hat{\imath}\,\mathrm{m^{2}}) \)
\( \vec{m} = -4\,\hat{\imath}\,\mathrm{A\,m^{2}} \)
\( \displaystyle U = -\vec{m}\cdot\vec{B} = -(-4\,\hat{\imath}\,\mathrm{A\,m^{2}})\cdot(0.5\,\hat{\imath}\,\mathrm{T}) \)
\( U = +2\,\mathrm{J} \)
The moment points opposite to the field, so the coil is in the position of maximum potential energy (unstable equilibrium).