An electron enters a chamber in which a uniform magnetic field is present as shown below. An electric field of appropriate magnitude is also applied, so that the electron travels undeviated without any change in its speed through the chamber. We are ignoring gravity. Then, the direction of the electric field is
opposite to the direction of the magnetic field
opposite to the direction of the electron's motion
normal to the plane of the paper and coming out of the plane of the paper
normal to the plane of the paper and into the plane of the paper
A particle having a mass of $10^{- 2} \,kg$ carries a charge of $5 \times 10^{-8}\, C.$ The particle is given an initial horizontal velocity of $10^5\, m/s $ in the presence of electric field $E$ and magnetic field $B.$ To keep the particle moving in a horizontal direction, it is necessary that
A proton of mass $1.67 \times {10^{ - 27}}\,kg$ and charge $1.6 \times {10^{ - 19}}\,C$ is projected with a speed of $2 \times {10^6}\,m/s$ at an angle of $60^\circ $ to the $X - $ axis. If a uniform magnetic field of $0.104$ $Tesla$ is applied along $Y - $ axis, the path of proton is
A deutron of kinetic energy $50\, keV$ is describing a circular orbit of radius $0.5$ $metre$ in a plane perpendicular to magnetic field $\overrightarrow B $. The kinetic energy of the proton that describes a circular orbit of radius $0.5$ $metre$ in the same plane with the same $\overrightarrow B $ is........$keV$
An electron is moving along the positive $X$$-$axis. You want to apply a magnetic field for a short time so that the electron may reverse its direction and move parallel to the negative $X$$-$axis. This can be done by applying the magnetic field along
A long solenoid has $100\,turns/m$ and carries current $i.$ An electron moves with in the solenoid in a circle of radius $2·30\,cm$ perpendicular to the solenoid axis. The speed of the electron is $0·046\,c$ ($c =$ speed of light). Find the current $i$ in the solenoid (approximate).....$A$