A beam of electrons is moving with constant velocity in a region having electric and magnetic fields of strength $ 20 V m^{-1} $ and 0.5 T at right angles to the direction of motion of the electrons. What is the velocity of the electrons?

Question

A long straight current-carrying wire is placed in a uniform magnetic field of strength $ B = 0.5 \, \text{T} $. If the current in the wire is $ I = 2 \, \text{A} $ and the wire makes an angle of $ 30^\circ $ with the magnetic field, find the force per unit length on the wire.

Answer 1

To determine the velocity of the electrons in the given setup where both electric and magnetic fields are present, we need to use the concept of velocity selector. In a velocity selector, only particles with a specific velocity are able to pass through without being deflected by the electric or magnetic fields.

The electric field ($E$) and magnetic field ($B$) are at right angles to the direction of motion of the electron beam. In such a setup, the condition for no deflection is that the magnetic force and electric force balance each other. The forces can be expressed as:

Electric Force: F_e = e \cdot E
Magnetic Force: F_m = e \cdot v \cdot B

For the beam to remain undeflected, these forces must be equal, so:

e \cdot E = e \cdot v \cdot B

By cancelling the charge ($e$) on both sides, we get:

E = v \cdot B

The electron velocity ($v$) can then be calculated using this equation:

v = \frac{E}{B}

Substitute the provided values for the electric field ($E = 20 \, \text{V m}^{-1}$) and the magnetic field ($B = 0.5 \, \text{T}$):

v = \frac{20 \, \text{V m}^{-1}}{0.5 \, \text{T}} = 40 \, \text{m s}^{-1}

Therefore, the correct velocity of the electrons, given that they remain undeflected, is 40 \, \text{m s}^{-1}. This matches the correct answer option.

Answer 2