Question

An electron moves in a circular trajectory with radius \(r_{\mathrm{i}}\) in a constant magnetic field. What is the final radius of the trajectory when the magnetic field is doubled? a) \(\frac{r_{i}}{4}\) b) \(\frac{r_{i}}{2}\) c) \(r_{\mathrm{i}}\) d) \(2 r_{\mathrm{i}}\) e) \(4 r_{\mathrm{i}}\)

Short answer

Answer: (b) The final radius is half the initial radius.

Step-by-step solution

Consider the centripetal force equation

Centripetal force, which is required to keep a particle moving in a circular path, can be described as: $$F_c = \dfrac{m v^2}{r}$$ where \(F_c\) is the centripetal force, \(m\) is the mass of the particle, \(v\) is the speed of the particle, and \(r\) is the radius of the circular path.

Relate the magnetic force to the centripetal force

The magnetic force acting on the electron can be described by the Lorentz force equation: $$F_m = qvB$$ where \(F_m\) is the magnetic force, \(q\) is the charge of the electron, and \(B\) is the magnetic field strength. Since the magnetic force is responsible for the centripetal motion of the electron, we can equate the magnetic force with the centripetal force: $$qvB = \dfrac{m v^2}{r}$$

Solve for the initial radius

Let's express the initial radius \(r_i\) in terms of the initial magnetic field, \(B_i\), the mass of the electron, \(m\), the charge of the electron, \(q\), and its speed, \(v\). From step 2, we have: $$qvB_i = \dfrac{m v^2}{r_i}$$ Solving for \(r_i\), we get: $$r_i = \dfrac{mv}{q B_i}$$

Compute the final radius

When the magnetic field is doubled, the final magnetic field \(B_f = 2B_i\). Now, the equation relating the variables in the final state is given by: $$qvB_f = \dfrac{m v^2}{r_f}$$ Where \(r_f\) is the final radius. We can now substitute \(B_f = 2B_i\) and rearrange the equation to find \(r_f\): $$r_f = \dfrac{mv}{q (2B_i)}$$

Compare the initial and final radii

Finally, let's compare the initial and final radii: $$\dfrac{r_f}{r_i} = \dfrac{\dfrac{mv}{q (2B_i)}}{\dfrac{mv}{q B_i}}$$ Simplify to find the ratio: $$\dfrac{r_f}{r_i} = \dfrac{1}{2}$$ Hence, the final radius of the trajectory when the magnetic field is doubled is half the initial radius: $$r_f = \dfrac{r_i}{2}$$ The correct answer is (b).