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Topic: Force on moving charge in magnetic and electric fields· Magnetic Effects of Current and Magnetism

An electron (mass 9×10^-31 kg and charge 1.6×10^-19C) 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 an uniform electric field E toget…

An electron (mass \(9\times10^{-31}\) kg and charge \(1.6\times10^{-19}\)C) moving with speed \(c/100\) (\(c\) = speed of light) is injected into a magnetic field \(\vec B\) of magnitude \(9\times10^{-4}\) T perpendicular to its direction of motion. We wish to apply an uniform electric field \(\vec E\) together with the magnetic field so that the electron does not deflect from its path. Then (speed of light \(c = 3\times10^{8}\) ms\(^{-1}\))

1. \(\vec E\) is perpendicular to \(\vec B\) and its magnitude is \(27\times10^{4}\) V m\(^{-1}\)

2. \(\vec E\) is perpendicular to \(\vec B\) and its magnitude is \(27\times10^{2}\) V m\(^{-1}\)

3. \(\vec E\) is parallel to \(\vec B\) and its magnitude is \(27\times10^{2}\) V m\(^{-1}\)

4. \(\vec E\) is parallel to \(\vec B\) and its magnitude is \(27\times10^{4}\) V m\(^{-1}\)

1 answer

Best answer

Answer: 2. \(\vec E\) is perpendicular to \(\vec B\) and its magnitude is \(27\times10^{2}\) V m\(^{-1}\)

Key idea: for the electron to go undeflected, the electric force must cancel the magnetic force, so \(qE=qvB\) with \(\vec E\perp\vec B\) (and both perpendicular to \(\vec v\)).

Speed \(v=c/100=3\times10^{6}\) m/s.

$$E=vB=3\times10^{6}\times 9\times10^{-4}=2.7\times10^{3}=27\times10^{2}\ \mathrm{V\,m^{-1}}$$

If \(\vec E\) were parallel to \(\vec B\), the electric force would be along the field while the magnetic force is perpendicular to it, so they could never cancel. Options 3 and 4 are therefore wrong, and option 1 has the wrong magnitude.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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