A charge $q$ moves in a region where electric field and magnetic field both exist, then force on it is
A charge $q$ moves in a region where electric field and magnetic field both exist, then force on it is
- [AIPMT 2002]
- A
$q \overrightarrow{ E }+ q (\overrightarrow{ V } \times \overrightarrow{ B })$
- B
$q (\overrightarrow{ V } \times \overrightarrow{ B })$
- C
$q \overrightarrow{ E }+ q (\overrightarrow{ B } \times \overrightarrow{ V })$
- D
$q \overrightarrow{ B }+ q (\overrightarrow{ E } \times \overrightarrow{ V })$
Similar Questions
The figure shows a region of length $'l'$ with a uniform magnetic field of $0.3\, T$ in it and a proton entering the region with velocity $4 \times 10^{5}\, ms ^{-1}$ making an angle $60^{\circ}$ with the field. If the proton completes $10$ revolution by the time it cross the region shown, $l$ is close to....... $m$
(mass of proton $=1.67 \times 10^{-27} \,kg ,$ charge of the proton $\left.=1.6 \times 10^{-19}\, C \right)$
The figure shows a region of length $'l'$ with a uniform magnetic field of $0.3\, T$ in it and a proton entering the region with velocity $4 \times 10^{5}\, ms ^{-1}$ making an angle $60^{\circ}$ with the field. If the proton completes $10$ revolution by the time it cross the region shown, $l$ is close to....... $m$
(mass of proton $=1.67 \times 10^{-27} \,kg ,$ charge of the proton $\left.=1.6 \times 10^{-19}\, C \right)$
- [JEE MAIN 2020]
An electron is moving along $+x$ direction with a velocity of $6 \times 10^{6}\, ms ^{-1}$. It enters a region of uniform electric field of $300 \,V / cm$ pointing along $+ y$ direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the $x$ direction will be
An electron is moving along $+x$ direction with a velocity of $6 \times 10^{6}\, ms ^{-1}$. It enters a region of uniform electric field of $300 \,V / cm$ pointing along $+ y$ direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the $x$ direction will be
- [JEE MAIN 2020]
Two parallel beams of protons and electrons, carrying equal currents are fixed at a separation $d$. The protons and electrons move in opposite directions. $P$ is a point on a line joining the beams, at a distance $x$ from any one beam. The magnetic field at $P$ is $B$. If $B$ is plotted against $x$, which of the following best represents the resulting curve
Two parallel beams of protons and electrons, carrying equal currents are fixed at a separation $d$. The protons and electrons move in opposite directions. $P$ is a point on a line joining the beams, at a distance $x$ from any one beam. The magnetic field at $P$ is $B$. If $B$ is plotted against $x$, which of the following best represents the resulting curve
A particle of mass $m$ carrying charge $q$ is accelerated by a potential difference $V$. It enters perpendicularly in a region of uniform magnetic field $B$ and executes circular arc of radius $R$, then $\frac{q}{m}$ equals
A particle of mass $m$ carrying charge $q$ is accelerated by a potential difference $V$. It enters perpendicularly in a region of uniform magnetic field $B$ and executes circular arc of radius $R$, then $\frac{q}{m}$ equals
Electron moves at right angles to a magnetic field of $1.5 \times 10^{-2}\,tesla$ with speed of $6 \times 10^7\,m/s$. If the specific charge of the electron is $1.7 \times 10^{11}\,C/kg$. The radius of circular path will be......$cm$
Electron moves at right angles to a magnetic field of $1.5 \times 10^{-2}\,tesla$ with speed of $6 \times 10^7\,m/s$. If the specific charge of the electron is $1.7 \times 10^{11}\,C/kg$. The radius of circular path will be......$cm$
- [AIIMS 2010]