Physics · Moving Charges and Magnetic Field

JEE Main 2026 — 21 January, Morning Shift — Question 32

A current carrying is placed vertically and a particle of mass m with charge Q is released from rest. The particle moves along the axis of solenoid. If g is acceleration due to gravity then the acceleration (a) of the charged particle will satisfy :

Option A:
$\mathrm{a}=\mathrm{g}$
Correct
Option B:
$\mathrm{a}>\mathrm{g}$
Option C:
$a=0$
Option D:
$0<$ a $<$ g

Answer: A

Step-by-step solution

Since the solenoid is placed vertically, the magnetic field inside the solenoid will be either along - y or +y axis. ⇒ Particle will gain velocity along -y axis. $\Rightarrow \quad \overrightarrow{\mathrm{F}}_{\mathrm{B}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})$ $\Rightarrow \quad \overrightarrow{\mathrm{F}}_{\mathrm{B}}=0$ $\Rightarrow \quad \overrightarrow{\mathrm{F}}_{\text {net }}=\mathrm{mg}$ $\Rightarrow \quad \mathrm{a}_{\text {net }}=\mathrm{g}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 21 January, Morning Shift
Subject: Physics
Chapter: Moving Charges and Magnetic Field
Topic: Motion of Charged Particles in magnetic Fields