Q. No.A particle of mass \(m\), charge \(Q\), and kinetic energy \(T\) enters a transverse uniform magnetic field of induction \(\vec B\). What will be the kinetic energy of the particle after seconds?
| 1. | \(3~\text{T}\) | 2. | \(2~\text{T}\) |
| 3. | \(\text{T}\) | 4. | \(4~\text{T}\) |
What is the result of an electric charge in uniform motion?
| 1. | an electric field only. |
| 2. | a magnetic field only. |
| 3. | both electric and magnetic field. |
| 4. | neither electric nor magnetic field. |
A particle of charge +q and mass m moving under the influence of a uniform electric field and a uniform magnetic field follows a trajectory from P to Q as shown in the figure. The velocities at P and Q are and respectively. Which of the following statement(s) is/are correct?
| 1. | \(\mathrm{E}=\frac{3}{4} \frac{\mathrm{mv}^2}{\mathrm{qa}}\) |
| 2. | Rate of work done by electric field at P is\(\frac{3}{4} \frac{\mathrm{mv}^3}{\mathrm{a}}\) |
| 3. | Rate of work done by both fields at Q is zero |
| 4. | All of the above |
| 1. | kinetic energy changes |
| 2. | kinetic energy remains constant |
| 3. | speed changes |
| 4. | momentum remains constant |
Moving perpendicular to field \(B\), a proton and an alpha particle both enter an area of uniform magnetic field \(B\). If the kinetic energy of the proton is \(1~\text{MeV}\) and the radius of the circular orbits for both particles is equal, the energy of the alpha particle will be:
1. \(4~\text{MeV}\)
2. \(0.5~\text{MeV}\)
3. \(1.5~\text{MeV}\)
4. \(1~\text{MeV}\)
A beam of electrons passes un-deflected through mutually perpendicular electric and magnetic fields. Where do the electrons move if the electric field is switched off and the same magnetic field is maintained?
| 1. | in an elliptical orbit. |
| 2. | in a circular orbit. |
| 3. | along a parabolic path. |
| 4. | along a straight line. |
| 1. | 8 N in - z-direction. |
| 2. | 4 N in the z-direction. |
| 3. | 8 N in the y-direction. |
| 4. | 8 N in the z-direction. |
A current-carrying wire is placed in a uniform magnetic field in the shape of the curve \(y= \alpha sin ({\pi x \over L}),~0 \underline{<}x \underline{<}~2L\)
. What will be the force acting on the wire?
| 1. | \(iBL \over \pi\) | 2. | \(iBL \pi\) |
| 3. | \(2iBL \) | 4. | Zero |
An electron is moving in a circular path under the influence of a transverse magnetic field of \(3.57\times 10^{-2}~\text{T}\). If the value of \(\frac{e}{m}\) is \(1.76\times 10^{11}~\text{C/kg}\), what will be the frequency of revolution of the electron?
| 1. | \(1~\text{GHz}\) | 2. | \(100~\text{MHz}\) |
| 3. | \(62.8~\text{MHz}\) | 4. | \(6.28~\text{MHz}\) |
| Statement I: | The electric force changes the speed of the charged particle and hence changes its kinetic energy: whereas the magnetic force does not change the kinetic energy of the charged particle. |
| Statement II: | The electric force accelerates the positively charged particle perpendicular to the direction of electric field. The magnetic force accelerates the moving charged particle along the direction of magnetic field. |
| 1. | Both Statement I and Statement II are correct. |
| 2. | Both Statement I and Statement II are incorrect. |
| 3. | Statement I is correct and Statement II is incorrect. |
| 4. | Statement I is incorrect and Statement II is correct. |
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