Question:
An electron is moving along +x direction with a velocity of \(6 \times 10^{6}\ \mathrm{ms^{-1}}\). It enters a region of uniform electric field of 300 V/cm pointing along +y direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the x direction will be :
An electron is moving along +x direction with a velocity of \(6 \times 10^{6}\ \mathrm{ms^{-1}}\). It enters a region of uniform electric field of 300 V/cm pointing along +y direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the x direction will be :
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Net force must be zero: \(eE = evB\) in magnitude, and the magnetic force on the electron must point along +y (opposite to the electric force on the electron).
Updated On: Nov 5, 2026
- \(3 \times 10^{-4}\) T, along \(-z\) direction
- \(5 \times 10^{-3}\) T, along \(-z\) direction
- \(5 \times 10^{-3}\) T, along \(+z\) direction
- \(3 \times 10^{-4}\) T, along \(+z\) direction
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The Correct Option is C
Solution and Explanation
E = 300 V/cm = \(3\times 10^{4}\) V/m along +y. Force on the electron (charge \(-e\)) is along \(-y\), magnitude eE.
For zero deflection the magnetic force must be along +y with the same magnitude: \(evB = eE\Rightarrow B = \frac{E}{v} = \frac{3\times 10^{4}}{6\times 10^{6}} = 5\times 10^{-3}\) T.
Direction: \(\vec F_B = -e(\vec v\times\vec B)\) must be along +y, so \(\vec v\times\vec B\) is along \(-y\). Since \(\hat i\times\hat k = -\hat j\), \(\vec B\) is along +z.
Answer: (3) \(5 \times 10^{-3}\) T, along +z direction
For zero deflection the magnetic force must be along +y with the same magnitude: \(evB = eE\Rightarrow B = \frac{E}{v} = \frac{3\times 10^{4}}{6\times 10^{6}} = 5\times 10^{-3}\) T.
Direction: \(\vec F_B = -e(\vec v\times\vec B)\) must be along +y, so \(\vec v\times\vec B\) is along \(-y\). Since \(\hat i\times\hat k = -\hat j\), \(\vec B\) is along +z.
Answer: (3) \(5 \times 10^{-3}\) T, along +z direction
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