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.