Question

A uniform electric field of magnitude E= 100kV/m is directed upward. Perpendicular to E and directed into the page, there exists a uniform magnetic field of magnitude B=0.5T. A beam of particles of charge +q enters this region. What should be the chosen speed of the particles for which the particles will not be deflected by the crossed electric and magnetic field? Assume that the particles are moving in +ve direction.

Answer 1

Hint: We will apply the formula of force in both the cases i.e. the magnetic and the electric field. Applying the right-hand screw rule, we will equate both the values of force in order to find the desired velocity. Refer to the solution below.

Formula used: $F = qE$, $F = qvB$.

Complete Step-by-Step solution:

The magnitude of the uniform electric field going into upwards direction- $E = 100kV/m$
The magnitude of the uniform magnetic field going into upwards direction- $B = 0.5T$
Since the charge +q enters this arrangement, there will be some force acting on it.
As we know that the formula of force due to electric field is $F = qE$. (direction – upwards)
And the formula of force due to magnetic field is $F = qvB$. (direction – downwards, applying the right-hand screw rule)
Since the two forces are equal and opposite, we can equate the formulas of the forces as well. We get-
$
   \Rightarrow qE = qvB \\
    \\
   \Rightarrow v = \dfrac{E}{B} \\
$ (cancelling the charge on both sides)
$
   \Rightarrow v = \dfrac{{100 \times 1000}}{{0.5}} \\
    \\
   \Rightarrow v = 2 \times {10^5}m/s \\
$
Hence, the particulate speed selected for which the particles are not deflected by the electrical and magnetic fields crossed is $2 \times {10^5}m/s$.

Note: The rule for the right-hand screw can be used when determining a direction based on a direction of rotation. The axis is 'grabbed' in the right hand, the fingers curl in the positive rotation and the thumb is bent in the positive direction.