Describe the features of the magnetic force acting on a charged particle moving inside a magnetic field.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The magnetic force $\overrightarrow{F_{m}}$ acting on a charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\overrightarrow{B}$ is given by:
$\overrightarrow{F_{m}} = q(\vec{v} \times \overrightarrow{B})$
$\therefore F_{m} = q v B \sin \theta$,where $\theta$ is the angle between $\vec{v}$ and $\overrightarrow{B}$.
Features:
$(i)$ The force depends on the charge $q$,velocity $\vec{v}$,and magnetic field $\overrightarrow{B}$. The force on a negative charge is opposite to that on a positive charge.
$(ii)$ The force is a vector product of velocity and magnetic field. If $\theta = 0^{\circ}$ or $\theta = 180^{\circ}$,then $F_{m} = q v B \sin(0^{\circ}) = 0$ or $F_{m} = q v B \sin(180^{\circ}) = 0$. The force acts in a direction perpendicular to both the velocity and the magnetic field,determined by the right-hand rule.
$(iii)$ The magnetic force is zero if the charge is stationary $(v = 0)$. Thus,only a moving charge experiences a magnetic force.

Explore More

Similar Questions

$A$ horizontal conductor is oriented north-south and carries some current. $A$ positively charged particle located vertically above it and having a velocity directed northward experiences an upward force. What is the direction of the force if this charged particle were located to the east of the conductor and had a velocity directed towards the conductor?

Assertion : $A$ proton and an alpha particle having the same kinetic energy are moving in circular paths in a uniform magnetic field. The radii of their circular paths will be equal.
Reason : Any two charged particles having equal kinetic energies and entering a region of uniform magnetic field $\overrightarrow{B}$ in a direction perpendicular to $\overrightarrow{B}$,will describe circular trajectories of equal radii.

$A$ charge having $q/m$ equal to $10^8 \, C/kg$ and with velocity $3 \times 10^5 \, m/s$ enters into a uniform magnetic field $0.3 \, T$ at an angle $30^{\circ}$ with the direction of the field. The radius of curvature will be ...... $cm$.

$A$ charged particle is moving along a magnetic field line. What is the magnetic force acting on the particle? $(\sin 0^{\circ}=0, \sin \frac{\pi}{2}=1)$

$A$ proton beam enters a magnetic field of $ 10^{-4} \,Wb m^{-2} $ normally. If the specific charge of the proton is $ 10^{11} \,C kg^{-1} $ and its velocity is $ 10^{9} \,ms^{-1} $, then the radius of the circle described will be (in $\,m$)

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo