Question

An electron (mass \(9 \times 10^{-31}\) kg and charge \(1.6 \times 10^{-19}\) C) moving with speed \(c/100\) (\(c\) = speed of light) is injected into a magnetic field of magnitude \(9 \times 10^{-4}\) T perpendicular to its direction of motion. We wish to apply a uniform electric field \( \vec{E} \) together with the magnetic field so that the electron does not deflect from its path. (speed of light \(c = 3 \times 10^8\) m/s):

• A. \( \vec{E} \) is perpendicular to \( \vec{B} \) and its magnitude is \( 2.7 \times 10^2 \) V m\(^{-1} \)
• B. \( \vec{E} \) is parallel to \( \vec{B} \) and its magnitude is \( 2.7 \times 10^2 \) V m\(^{-1} \)
• C. \( \vec{E} \) is parallel to \( \vec{B} \) and its magnitude is \( 2.7 \times 10^6 \) V m\(^{-1} \)
• D. \( \vec{E} \) is perpendicular to \( \vec{B} \) and its magnitude is \( 2.7 \times 10^6 \) V m\(^{-1} \)

Hint:

For an undeflected motion of a charged particle in crossed electric and magnetic fields (\( \vec{E} \perp \vec{B} \)), the condition is \( \vec{E} = - (\vec{v} \times \vec{B} ) \), and the magnitude is \( E = v B \) when \( \vec{v} \perp \vec{B} \). The electric field must be perpendicular to both velocity and magnetic field.

Answer 1

The Correct Option is A

To resolve the issue, we must establish the magnitude and orientation of the electric field \( \vec{E} \) required to counteract the electron's deflection when subjected to a magnetic field \( \vec{B} \). When no deflection is observed, the magnetic force and the electric force are equal in magnitude.

Step 1: Calculate the magnetic force (\( F_B \)) on the electron.

The magnetic force is quantified by the equation:

\( F_B = qvB \sin\theta \)

Given that \( \vec{B} \) is perpendicular to the electron's velocity (\( \theta = 90^\circ \)), \( \sin\theta \) equals 1. Consequently:

\( F_B = qvB \)

Input the provided values:

• Electron charge (\( q \)) = \( 1.6 \times 10^{-19} \) C
• Electron speed (\( v \)) = \( \frac{c}{100} = \frac{3 \times 10^8}{100} = 3 \times 10^6 \) m/s
• Magnetic field (\( B \)) = \( 9 \times 10^{-4} \) T

Perform the substitution:

\( F_B = (1.6 \times 10^{-19})(3 \times 10^6)(9 \times 10^{-4}) \)

\( F_B = 4.32 \times 10^{-16} \) N

Step 2: Determine the electric field (\( \vec{E} \)) necessary to balance this force.

To prevent deflection, the electric force \( F_E \) must be equivalent to \( F_B \):

\( F_E = qE = F_B \)

\( E = \frac{F_B}{q} \)

Substitute the values:

\( E = \frac{4.32 \times 10^{-16}}{1.6 \times 10^{-19}} \)

\( E = 2.7 \times 10^2 \) V/m

Conclusion:

For the electron to remain undeflected, the electric field \( \vec{E} \) must be perpendicular to \( \vec{B} \) and possess a magnitude of \( 2.7 \times 10^2 \) V/m. Therefore, the accurate statement is: \( \vec{E} \) is perpendicular to \( \vec{B} \) and its magnitude is \( 2.7 \times 10^2 \) V/m.