An electron (mass 9×10^-31 kg and charge 1.6×10^-19 C) moving with speed c/100 (c= speed of light) is injected into a magnetic fi…

Question

An electron (mass 9×10^-31 kg and charge 1.6×10^-19 C) moving with speed c/100 (c= speed of light) is injected into a magnetic field B of magnitude 9×10^-4 T perpendicular to its direction of motion. We wish to apply a uniform electric field E together with the magnetic field so that the electron does not deflect from its path. Then (speed of light c=3×10^8 m s^-1):

1. A. E is perpendicular to B and its magnitude is 27×10^4 V m^-1
2. B. E is perpendicular to B and its magnitude is 27×10^2 V m^-1 (Correct answer)
3. C. E is parallel to B and its magnitude is 27×10^2 V m^-1
4. D. E is parallel to B and its magnitude is 27×10^4 V m^-1

Correct Answer

Option B — E is perpendicular to B and its magnitude is 27×10^2 V m^-1

Detailed Solution & Explanation

The correct answer is Option B.

Key Points

• For zero deflection the electric force must balance the magnetic force, which requires $\vec{E}=\vec{v}\times\vec{B}$, so $\vec{E}\perp\vec{B}$.
• Magnitude: $E=vB=\dfrac{c}{100}\times B=\dfrac{3\times10^{8}}{100}\times 9\times10^{-4}$.
• $E = 3\times10^{6}\times 9\times10^{-4}=27\times10^{2}$ V m$^{-1}$.