Question

A game of Electrons accelerated through a potential difference V It is in past normally through a uniform magnetic field where it moves in a circle of radius R it would have moved in a circle of radius two R if it were initially accelerated through a potential difference

• A. V
• B. 2V
• C. 3V
• D. 4V

Answer 1

Answer: (D)

4V

Answer 2

• Energy gained by electron: When an electron (charge e, mass m) is accelerated through a potential difference V, its potential energy is converted into kinetic energy.

  $$\frac{1}{2}mv^2 = eV$$

  From this, the velocity of the electron is $v = \sqrt{\frac{2eV}{m}}$.

• Motion in magnetic field: When this electron enters a uniform magnetic field B perpendicular to its velocity, it moves in a circular path. The magnetic force provides the necessary centripetal force:

  $$evB = \frac{mv^2}{R}$$

  Solving for the radius R:

  $$R = \frac{mv}{eB}$$

• Relating R and V: Substitute the expression for v from the first step into the equation for R:

  $$R = \frac{m}{eB} \sqrt{\frac{2eV}{m}}$$ $$R = \frac{1}{B} \sqrt{\frac{m^2 \cdot 2eV}{e^2 m}}$$ $$R = \frac{1}{B} \sqrt{\frac{2mV}{e}}$$

  From this equation, we can see that R is directly proportional to the square root of V, assuming m, e, and B are constant:

  $$R \propto \sqrt{V}$$

  This implies:

  $$V \propto R^2$$

• Calculating the new potential difference: Let the initial potential difference be V and the initial radius be R. Let the new potential difference be V' when the radius becomes 2R.

  Using the proportionality $V \propto R^2$:

  $$\frac{V'}{V} = \left(\frac{R_{new}}{R_{initial}}\right)^2$$

  Given $R_{initial} = R$ and $R_{new} = 2R$:

  $$\frac{V'}{V} = \left(\frac{2R}{R}\right)^2$$ $$\frac{V'}{V} = (2)^2$$ $$\frac{V'}{V} = 4$$ $$V' = 4V$$

The electron would have moved in a circle of radius two R if it were initially accelerated through a potential difference of 4V.