JEE Main 202623 JanuaryMorning Shift
- Exam
- JEE Main
- Year
- 2026
- Date
- 23 January
- Shift
- Morning Shift
- Subject
- Physics
- Chapter
- Electromagnetic Induction
JEE Main 2026 · 23 January · Morning Shift · Physics · Electromagnetic Induction
A simple pendulum made of mass 10 g and a metallic wire of length...
A simple pendulum made of mass 10 g and a metallic wire of length 10 cm is suspended vertically in a uniform magnetic field of 2 T. The magnetic field direction is perpendicular to the plane of oscillations of the pendulum. If the pendulum is released from an angle of with vertical, then maximum induced EMF between the point of suspension and point of oscillation is mV. (Take )
Correct answer
100
Step-by-step solution
The maximum induced EMF for a rod of length rotating in a magnetic field is:
\varepsilon_\max = \frac{B \omega_\max \ell^2}{2} \quad \cdots(1)
Using energy conservation to find \omega_\max:
From equation (1):
\varepsilon_\max = \frac{2 \times 10 \times 0.01}{2} = 0.1 \text{ V} = 100 \text{ mV}
\varepsilon_\max = \frac{B \omega_\max \ell^2}{2} \quad \cdots(1)
Using energy conservation to find \omega_\max:
From equation (1):
\varepsilon_\max = \frac{2 \times 10 \times 0.01}{2} = 0.1 \text{ V} = 100 \text{ mV}