Question

When an electron placed in a uniform magnetic field is accelerated from rest through a potential difference \( V_1 \), it experiences a force \( F \). If the potential difference is changed to \( V_2 \), the force experienced by the electron in the same magnetic field is \( 2F \), then the ratio of potential differences \( \frac{V_2}{V_1} \) is:

• A. \( 2 : 1 \)
• B. \( 1 : 4 \)
• C. \( 4 : 1 \)
• D. \( 1 : 2 \)

Hint:

For problems involving force and potential difference in a magnetic field, remember that the force is proportional to the velocity of the electron, which is related to the square root of the potential difference.

Answer 1

The Correct Option is C

To determine the ratio of potential differences \( \frac{V_2}{V_1} \), we examine the relationship between the force experienced by an electron in a magnetic field and its velocity. The force \( F \) experienced by an electron moving with velocity \( v \) perpendicular to a magnetic field \( B \) is given by the equation:

\( F = e v B \)

where \( e \) is the charge of the electron.

When the electron is accelerated from rest through a potential difference \( V \), it gains kinetic energy equal to the work done by the electric field, which is:

\( eV = \frac{1}{2}mv^2 \)

where \( m \) is the mass of the electron.

Solving for \( v \), we have:

\( v = \sqrt{\frac{2eV}{m}} \)

Inserting this into the expression for the force, the force \( F \) becomes:

\( F = eB\sqrt{\frac{2eV}{m}} \)

Thus, \( F \propto \sqrt{V} \).

According to the problem, when the potential difference is \( V_2 \), the force experienced is \( 2F \). Therefore:

\( 2F = eB\sqrt{\frac{2eV_2}{m}} \)

Since \( F = eB\sqrt{\frac{2eV_1}{m}} \), we equate and solve for the ratio:

\( 2\sqrt{V_1} = \sqrt{V_2} \)

Squaring both sides gives:

\( 4V_1 = V_2 \)

Therefore, the ratio \( \frac{V_2}{V_1} = \frac{4}{1} \).

Hence, the correct answer is \( 4:1 \).

Answer 2

We are given that the force experienced by the electron is proportional to the velocity, and the velocity of the electron is related to the kinetic energy. The kinetic energy is given by the potential difference through which the electron is accelerated. The force is directly proportional to the velocity in a uniform magnetic field. 
Step 1: The kinetic energy acquired by the electron is given by: \[ KE = eV_1 \] where \( e \) is the charge of the electron, and \( V_1 \) is the initial potential difference. The velocity of the electron is: \[ v_1 = \sqrt{\frac{2eV_1}{m}}. \] The force \( F \) experienced by the electron in the magnetic field is given by: \[ F = evB. \] 
Step 2: For the second potential difference \( V_2 \), the velocity becomes: \[ v_2 = \sqrt{\frac{2eV_2}{m}}. \] The force experienced by the electron is: \[ F_2 = ev_2B = e\sqrt{\frac{2eV_2}{m}}B. \] We are given that \( F_2 = 2F \), so: \[ 2F = F_2 = ev_2B. \] 
Step 3: Now, using the ratio of forces: \[ \frac{F_2}{F} = \frac{v_2}{v_1} = \frac{\sqrt{\frac{2eV_2}{m}}}{\sqrt{\frac{2eV_1}{m}}} = \sqrt{\frac{V_2}{V_1}}. \] Since \( F_2 = 2F \), we have: \[ \sqrt{\frac{V_2}{V_1}} = 2. \] Squaring both sides gives: \[ \frac{V_2}{V_1} = 4. \] Thus, the ratio of potential differences is \( \frac{V_2}{V_1} = 4 : 1 \).