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 value of \( x \), we must first find the electric field magnitude in the region where a proton travels undeflected due to balanced crossed electric and magnetic fields.

The proton moves at a constant speed of \( 2 \times 10^5 \, \text{m/s} \) when the electric field \( \mathbf{E} \) and magnetic field \( \mathbf{B} \) are mutually perpendicular and of such strengths that their forces on the proton negate each other:

1. The electric force is \( F_E = qE \), where \( q \) is the proton's charge.

2. The magnetic force on the proton moving at speed \( v \) in field \( \mathbf{B} \) is \( F_B = qvB \), where \( B \) is the magnetic field strength.

3. For zero net force, \( F_E = F_B \).

Therefore, \( qE = qvB \), which simplifies to:

\( E = vB \)

Next, consider the scenario when the electric field is absent:

1. The proton follows a circular path, indicating the magnetic force acts as the centripetal force for a circular trajectory with radius \( r = 0.02 \, \text{m} \).

2. The centripetal force is \( F_c = \frac{mv^2}{r} \).

3. Equating the magnetic and centripetal forces yields \( qvB = \frac{mv^2}{r} \).

This equation simplifies to:

\( B = \frac{mv}{qr} \)

Substitute the known values: \( m = 1.6 \times 10^{-27} \, \text{kg} \), \( v = 2 \times 10^5 \, \text{m/s} \), \( q = 1.6 \times 10^{-19} \, \text{C} \), and \( r = 0.02 \, \text{m} \):

• \( B = \frac{1.6 \times 10^{-27} \times 2 \times 10^5}{1.6 \times 10^{-19} \times 0.02} \approx 0.01 \, \text{T} \)

Now, substitute \( B = 0.01 \, \text{T} \) and \( v = 2 \times 10^5 \, \text{m/s} \) into the equation for \( E \):

• \( E = 2 \times 10^5 \times 0.01 = 2000 \, \text{N/C} \)

The electric field magnitude is \( 2000 \, \text{N/C} \), equivalent to \( 2 \times 10^3 \, \text{N/C} \). Given the conversion \( x \times 10^4 \, \text{N/C} = E \):

• \( x \times 10^4 = 2000 \Rightarrow x = 0.2 \)

Thus, the value of \( x \) is 0.2.