JEE Main 2025 Apr 3 Shift 2, Physics Q1: Torque and Work Done in a Magnetic Field

JEE Main2025Apr 3, Shift 2Physics

A magnetic dipole experiences a torque of $803$ N m when placed in uniform magnetic field in such a way that dipole moment makes angle of $6 0^{\circ}$ with magnetic field. The potential energy of the dipole is :

A
$80$ J
B
$- 403$ J
C
$- 60$ J
D
$- 80$ J

Solution

For a magnetic dipole in a uniform field, $τ = M B sin θ$ and $U = - M B cos θ$ .

Using $θ = 6 0^{\circ}$ and $τ = 803$ :

$M B \cdot \frac{3}{2} = 803 \Rightarrow M B = 160$ J.

Therefore $U = - M B cos 6 0^{\circ} = - 160 \cdot \frac{1}{2} = - 80$ J.

Concept behind this question

Torque and Work Done in a Magnetic Field Notes, formulas and examples →

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Apr 3 Shift 2, Physics Q1: Torque and Work Done in a Magnetic Field

Concept behind this question

More previous year questions on Torque and Work Done in a Magnetic Field

Practice more Physics