Question

A moving charge will gain energy due to the application of
A. Electric field
B. Magnetic field
C. Both of these
D. None of these

Answer 1

Hint: A body gains energy when the net non-zero work is done on the body by external agents. Using the work-energy theorem, the gain in energy is equal to the work done. So, a moving charge will gain energy when the work is done on it.



Formula used:
\[\overrightarrow F = q\vec E + q\left( {\vec v \times \vec B} \right)\], here \[\vec F\] is the magnetic force vector,\[\vec E\] is the electric field, \[\vec v\] is the velocity of the charged particle and \[\vec B\] is the magnetic field in the region.



Complete answer:
When a charge is moving in a region with electric field then there must be potential difference in the region because \[E = - \dfrac{{dV}}{{dr}}\]
When a charge is moving through a region of potential difference, then the applied potential difference does work on charge. The magnitude of the work done by the potential difference is \[W = q\left( {\Delta V} \right)\].
Using the work-energy theorem, the moving charge will gain energy as net non-zero work is done on it.
When a moving charge is moving in a region of the magnetic field then the charge experiences magnetic force as \[\overrightarrow F = q\left( {\vec v \times \vec B} \right)\]
As the vector product is in the direction normal to the two vectors. So the magnetic force on the charge is perpendicular to the velocity of the charge, i.e. displacement of the charge.
So, the work done by the perpendicular force is zero. As there is no work done on the charge by the magnetic field, so there will be no gain in the energy of the moving charge.

Therefore, the correct option is (A).



Note:When the electric field and magnetic field are applied simultaneously then there is a possibility that the electric force will balance the magnetic force and hence there will be no work done on the charge and so the charge will not gain energy. That is why we didn’t choose the option of electric field and magnetic field applied.