$A$ beam of electrons passes undeflected through mutually perpendicular electric and magnetic fields. If the electric field is switched off,and the same magnetic field is maintained,the electrons move
- Ain an elliptical orbit
- Bin a circular orbit
- Calong a parabolic path
- Dalong a straight line
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$A$ proton of velocity $\vec{v} = (3 \hat{i} + 2 \hat{j}) \text{ ms}^{-1}$ enters a magnetic field of induction $\vec{B} = (2 \hat{j} + 3 \hat{k}) \text{ T}$. The acceleration produced in the proton in $\text{ms}^{-2}$ is (Specific charge of proton $= 0.96 \times 10^8 \text{ C kg}^{-1}$)
DifficultTS EAMCET 2002
View Solution$A$ charge $q_0$ moving with velocity $\overrightarrow{v}$ in a magnetic field of induction $\overrightarrow{B}$ experiences a force $\overrightarrow{F}$. The angle between $\overrightarrow{v}$ and $\overrightarrow{B}$ is $\theta$. The speed of $q_0$ after one second will be
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
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View Solution$A$ charged particle moves through a magnetic field in a direction perpendicular to it. Then the
$A$ proton, a deuteron, and an $\alpha$-particle with the same kinetic energy enter a uniform magnetic field at a right angle to the magnetic field. The ratio of the radii of their respective circular paths is
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