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 solve the problem of determining the magnitude of the electric field \( E \) when a proton is moving undeviated in crossed electric (\( E \)) and magnetic (\( B \)) fields, we need to understand the balance of forces involved. Initially when the proton is undeviated, the electric force must balance the magnetic force:

\( qE = qvB \)

Where:

• \( q \) is the charge of the proton (\( 1.6 \times 10^{-19} \, \text{C} \)).
• \( v \) is the velocity of the proton (\( 2 \times 10^5 \, \text{m/s} \)).

This simplifies to:

\( E = vB \)

When the electric field is turned off, and the proton moves in a circular path due to only the magnetic force, the centripetal force required for circular motion is provided by the magnetic force:

\( \frac{mv^2}{r} = qvB \)

Where:

• \( m \) is the mass of the proton (\( 1.6 \times 10^{-27} \, \text{kg} \)).
• \( r \) is the radius of the circular path (\( 0.02 \, \text{m} \)).

Rearranging gives:

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

Substituting the known values:

\( B = \frac{1.6 \times 10^{-27} \times 2 \times 10^5}{1.6 \times 10^{-19} \times 0.02} \)

\( B = \frac{3.2 \times 10^{-22}}{3.2 \times 10^{-21}} \)

\( B = 0.1 \, \text{T} \)

Substituting \( B \) back into \( E = vB \):

\( E = 2 \times 10^5 \times 0.1 \)

\( E = 2 \times 10^4 \, \text{N/C} \)

Now, the magnitude of the electric field is \( x \times 10^4 \, \text{N/C} \), where \( x = 2 \). According to the problem's expected range, \( x = 2 \) indeed falls within the given range of 1,10, verifying the solution.