Force to due to magnetic field
F=evB=Rmv2
with electric field also, net force is zero.
∴eE=evB
⇒E=vB
E=meBR.B
⇒m=EeB2R=1001.6×10−19×(0.5)2.(0.5×10−2)kg
=2.0×10−24kg
JEE Main 2019 — Physics Electromagnetism
A particle having the same charge as of electron moves in a circular path of radius 0.5cm under the influence of a magnetic field of 0.5T. If an electric field of 100 V/m makes it to move in a straight path, then the mass of the particle is (Given charge of electron =1.6×10−19C )
Held on 9 Jan 2019 · Verified 6 Jul 2026.
9.1×10−31kg
1.6×10−27kg
2.0×10−24kg
1.6×10−19kg
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
A magnetic field vector in an electromagnetic wave is represented by $\vec{B}=B_0\sin\left(2\pi vt-\dfrac{2\pi x}{\lambda}\right)\hat{j}$. Its associated electric field vector is ______.
A plane electromagnetic wave is moving in free space with velocity $c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ and its electric field is given as $\vec{E}=54 \sin (k z-\omega t) \hat{j} \mathrm{~V} / \mathrm{m}$, where $\hat{j}$ is the unit vector along $y$-axis. The magnetic field vector $\vec{B}$ of the wave is:
A charged particle moves in a uniform magnetic field. If the velocity is perpendicular to the field, the path of the particle is:
A monochromatic source of light operating at $15$ kW emits $2.5\times 10^{22}$ photons/s. The region of an electromagnetic spectrum to which the emitted electromagnetic radiation belongs to ________. (Take $h=6.6\times 10^{-34}$ J·s and $c=3\times 10^8$ m/s).
An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity, $\sigma=10 \mathrm{mho} / \mathrm{m}$. The ratio of maximum conduction current density to maximum displacement current density is $\_\_\_\_$. $\left[\text{Take }\frac{1}{4 \pi \epsilon_{\mathrm{o}}}=9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right]$
Work through every JEE Main Electromagnetism PYQ, year by year.