A circular coil having 200 turns, 2.5 × 10⁻⁴ m² area and carrying 100 current is placed in a uniform magnetic field of 1 T. Initially the magnetic dipole moment (M) was directed along B. Amount of work, required to rotate the coil through 90° from its initial orientation such that M becomes perpendicular to B, is ________ μ J. Numerical answer type. Work it out, then open the answer.

Answer: 5. W = mB(cos 0° - cos 90°) = NIAB W = 200 × 100×10⁻⁶ × 2.5×10⁻⁴ × 1 = 5×10⁻⁶ J So W = 5 μ J.

A circular coil having 200 turns, 2.5 × 10⁻⁴ m² area and carrying 100… | JEE Main 2024 (6 Apr, Shift 1)

Q39JEE Main 2024 · 6 APR · SHIFT 1Medium

Current Loop as a Magnetic Dipole

A circular coil having 200 turns, $2.5 \times 10^{-4}\text{ m}^{2}$ area and carrying $100\ \mu\text{A}$ current is placed in a uniform magnetic field of $1\text{ T}$ . Initially the magnetic dipole moment $(\vec{M})$ was directed along $\vec{B}$ . Amount of work, required to rotate the coil through $90^\circ$ from its initial orientation such that $\vec{M}$ becomes perpendicular to $\vec{B}$ , is ________ $\mu$ J.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 5

$W = mB(\cos0^\circ - \cos90^\circ) = NIAB$
$W = 200 \times 100\times10^{-6} \times 2.5\times10^{-4} \times 1 = 5\times10^{-6}\text{ J}$
So $W = 5\ \mu$ J.

Report a mistake

Current Loop as a Magnetic Dipole question from JEE Main 2024 (6 Apr, Shift 1)

More questions on Current Loop as a Magnetic Dipole