Question

A beam of well collimated cathode rays traveling with a speed of \[5 \times {10^6}m{s^{ - 1}}\]enter a region of mutually perpendicular electric and magnetic fields and emerge undeviated from this region. If\[\left| B \right| = 0.02T\], the magnitude of the electric field is
A. \[{10^5}V{m^{ - 1}}\]
B. \[2.5 \times {10^8}V{m^{ - 1}}\]
C. \[1.25 \times {10^{10}}V{m^{ - 1}}\]
D. \[2 \times {10^3}V{m^{ - 1}}\]

Answer 1

Hint: For undeviated motion, the net acceleration of the particle should be zero and hence using Newton’s 2nd law of motion, the net force acting on the particle is zero. We use Lorentz's force law as the particle is moving in an electric field and magnetic field.


Formula used:
\[\overrightarrow F = q\overrightarrow E + q\left( {\overrightarrow v \times \overrightarrow B } \right)\], here F is the net force acting on the particle of charge q moving with speed v in a region with magnetic field B and electric field E.



Complete answer:
The speed of the beam of cathode rays is \[5 \times {10^6}m{s^{ - 1}}\]
The magnetic field strength in the given region is given as 0.02T
It is given that the velocity, electric field and the magnetic field are mutually perpendicular.
For the undeviated motion of the cathode rays, the net force acting on the cathode rays must be equal to zero.
\[\overrightarrow F = 0\]
\[q\overrightarrow E + q\left( {\overrightarrow v \times \overrightarrow B } \right) = 0\]
When the motion is undeviated then the magnitude of the electric field is independent of the charge of the particle which is in motion in the region where the magnetic field and the electric field are present simultaneously.
\[\left| E \right| = vB\]
\[\left| E \right| = 5 \times {10^6} \times 0.02V{m^{ - 1}}\]
\[\left| E \right| = {10^5}V{m^{ - 1}}\]
So, the magnitude of the electric field in the region is \[{10^5}V{m^{ - 1}}\]
Therefore, the correct option is (A).


Note:We should be careful about the vector sum of the forces being zero. As the direction in the question is not specified, the vector sum of the forces will have only one component which should be equal to zero for the undeviate motion of the cathode rays.