Question

A proton is moving undeflected in a region of crossed electric and magnetic fields at a constant speed of \( 2 \times 10^5 \, \text{m/s} \). When the electric field is switched off, the proton moves along a circular path of radius 2 cm. The magnitude of electric field is \( x \times 10^4 \, \text{N/C} \). The value of \( x \) is \(\_\_\_\_\_\). (Take the mass of the proton as \( 1.6 \times 10^{-27} \, \text{kg} \)).

Hint:

When a proton moves undeflected in crossed electric and magnetic fields, the forces due to the electric and magnetic fields are equal in magnitude and opposite in direction, allowing you to solve for the electric field and magnetic field.

Answer 1

Correct Answer: 1

In the crossed electric and magnetic fields, the proton experiences a force due to both fields that keeps it moving undeflected. The magnetic force \( F_B \) and electric force \( F_E \) balance each other. The force due to the magnetic field is given by: \[ F_B = q v B, \] where \( q \) is the charge of the proton, \( v \) is the speed, and \( B \) is the magnetic field. The electric force is: \[ F_E = q E, \] where \( E \) is the electric field. At equilibrium, \( F_B = F_E \), so: \[ q v B = q E \quad \Rightarrow \quad v B = E. \] The proton moves along a circular path due to the magnetic field, so the centripetal force is: \[ F_{\text{centripetal}} = \frac{m v^2}{r}. \] Equating this with the magnetic force \( F_B \), we get: \[ \frac{m v^2}{r} = q v B \quad \Rightarrow \quad B = \frac{m v}{q r}. \] Using the known values for the mass of the proton \( m = 1.6 \times 10^{-27} \, \text{kg} \), radius \( r = 0.02 \, \text{m} \), and speed \( v = 2 \times 10^5 \, \text{m/s} \), we can find the electric field \( E \) using the relationship \( E = v B \). Thus, the value of \( x \) is \( \boxed{1} \).