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

To determine the magnitude of the electric field \( E \) for an undeviated proton in crossed electric (\( E \)) and magnetic (\( B \)) fields, the electric force must equal the magnetic force: \( qE = qvB \). Here, \( q \) is the proton's charge (\( 1.6 \times 10^{-19} \, \text{C} \)) and \( v \) is its velocity (\( 2 \times 10^5 \, \text{m/s} \)). This simplifies to \( E = vB \).

When the electric field is removed and the proton moves circularly due to the magnetic force alone, the magnetic force provides the centripetal force: \( \frac{mv^2}{r} = qvB \). Here, \( m \) is the proton's mass (\( 1.6 \times 10^{-27} \, \text{kg} \)) and \( r \) is the circular path's radius (\( 0.02 \, \text{m} \)). Rearranging yields \( B = \frac{mv}{qr} \).

Substituting values: \( B = \frac{1.6 \times 10^{-27} \times 2 \times 10^5}{1.6 \times 10^{-19} \times 0.02} = \frac{3.2 \times 10^{-22}}{3.2 \times 10^{-21}} = 0.1 \, \text{T} \).

Substituting \( B \) into \( E = vB \): \( E = 2 \times 10^5 \times 0.1 = 2 \times 10^4 \, \text{N/C} \).

The electric field magnitude is \( x \times 10^4 \, \text{N/C} \), with \( x = 2 \). This value falls within the specified range of 1 to 10.