Question

A proton is moving under the combined influence of an electric field \((E=1000 . \mathrm{V} / \mathrm{m})\) and a magnetic field \((B=1.20 \mathrm{~T}),\) as shown in the figure. a) What is the acceleration of the proton at the instant it enters the crossed fields? b) What would the acceleration be if the direction of the proton's motion were reversed?

Short answer

Answer: For the initial motion, the acceleration of the proton is a = (0, 9.58 × 10^7, 0) m/s^2. When the proton's motion is reversed along the x-axis, the acceleration is a' = (0, 9.58 × 10^7, -qv_0B_0/m) m/s^2, where q is the charge of the proton, v_0 is the initial velocity of the proton, B_0 is the magnetic field, and m is the mass of the proton.

Step-by-step solution

Calculate the initial Lorentz force

First, we need to calculate the Lorentz force acting on the proton. The formula for the Lorentz force is F = q(E + v × B). Since the proton is moving along the x-axis, we have v = (v_0, 0, 0), where v_0 is the initial velocity of the proton. The electric and magnetic fields are given as E = (0, E_0, 0) and B = (0, 0, B_0). So, the vector product v × B = (0, 0, v_0B_0) and the sum E + v × B = (0, E_0, v_0B_0). Now, we find the Lorentz force in the following manner: F = q(0, E_0, v_0B_0) = (0, qE_0, qv_0B_0)

Calculate the acceleration of the proton

Now that we have the force acting on the proton, we can calculate the acceleration using Newton's second law: F = m*a Since the force is a vector, the acceleration is also a vector: a = F/m = (0, qE_0/m, qv_0B_0/m) Using the given values, we can calculate the numerical value of the acceleration in the y-direction: a_y = qE_0/m = (1.6 × 10^-19 C)(1000 V/m) / (1.67 × 10^-27 kg) ≈ 9.58 × 10^7 m/s^2 The acceleration of the proton in the x and z direction is 0 m/s^2. So, the acceleration of the proton in the crossed fields is a = (0, 9.58 × 10^7, 0) m/s^2.

Calculate the acceleration of the proton with reversed motion

Now, let's imagine the proton's velocity reverses so the velocity is now -v_0 in the x-direction. We can follow the same steps as before, but this time with -v_0 as the x-component of velocity. The new Lorentz force will be: F' = q(0, E_0, -v_0B_0) = (0, qE_0, -qv_0B_0) The new acceleration of the proton is: a' = F'/m = (0, qE_0/m, -qv_0B_0/m) Using the same values as before, the acceleration in the y-direction remains the same: a'_y = qE_0/m ≈ 9.58 × 10^7 m/s^2 The acceleration in the z direction will change sign, a'_z = -qv_0B_0/m So, the acceleration of the proton with reversed motion is a' = (0, 9.58 × 10^7, -qv_0B_0/m) m/s^2.