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 \(E\hat i\) and a uniform magnetic field \(\mathrm B\hat k\) follows a trajectory from \(P\) to \(Q\) as shown in the figure. The velocities at \(P\) and \(Q\) are \(v\hat i\) and \(-2v\hat j\) respectively. Which of the following statement(s) is/are correct?
1. | \(E=\frac{3}{4} \frac{{mv}^2}{{qa}}\). |
2. | Rate of work done by electric field at \(P\) is \(\frac{3}{4} \frac{{mv}^3}{a}\). |
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 \(-z\text-\)direction. |
2. | \(4\) N in the \(z\text-\)direction. |
3. | \(8\) N in the \(y\text-\)direction. |
4. | \(8\) N in the \(z\text-\)direction. |
A current-carrying wire is placed in a uniform magnetic field in the shape of the curve \(y= \alpha \sin \left({\pi x \over L}\right),~0 \le x \le2L.\)
What will be the force acting on the wire?
1. | \(iBL \over \pi\) | 2. | \(iBL \pi\) |
3. | \(2iBL \) | 4. | zero |
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. |