A metallic block carrying current $I$ is subjected to a uniform magnetic induction $\overrightarrow B $ as shown in the figure. The moving charges experience a force $\overrightarrow F $ given by ........... which results in the lowering of the potential of the face ........ Assume the speed of the carriers to be $v$
$eVB\,\hat k$, $ABCD$
$eVB\,\hat k$, $EFGH$
$ - eVB\,\hat k$, $ABCD$
$ - eVB\,\hat k$, $EFGH$
A uniform magnetic field $B$ is acting from south to north and is of magnitude $1.5$ $Wb/{m^2}$. If a proton having mass $ = 1.7 \times {10^{ - 27}}\,kg$ and charge $ = 1.6 \times {10^{ - 19}}\,C$ moves in this field vertically downwards with energy $5\, MeV$, then the force acting on it will be
The acceleration of an electron at a moment in a magentic field $\vec B\, = \,2\hat i + 3\hat j + 4\hat k$ is $\vec a\, = \,x\hat i - 2\hat j + \hat k$. The value of $x$ is
A particle of mass $m$ and charge $q$ enters a region of magnetic field (as shown) with speed $v$. There is a region in which the magnetic field is absent, as shown. The particle after entering the region collides elas tically with a rigid wall. Time after which the velocity of particle becomes anti parallel to its initial velocity is
An electron, a proton, a deuteron and an alpha particle, each having the same speed are in a region of constant magnetic field perpendicular to the direction of the velocities of the particles. The radius of the circular orbits of these particles are respectively $R_e, R_p, R_d \,$ and $\, R_\alpha$. It follows that
A current carrying long solenoid is placed on the ground with its axis vertical. A proton is falling along the axis of the solenoid with a velocity $v$. When the proton enters into the solenoid, it will