A charged particle with charge q enters a region of constant, uniform and mutually orthogonal fields E and B with a velocity v perpendicular to both E - Physics MCQ AIEEE 2007 2026 | ExamTest.live

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A charged particle with charge q enters a region of constant, uniform and mutually orthogonal fields E and B with a velocity v perpendicular to both E and B, and comes out without any change in magnitude or direction of v . Then:

Solution & Step-by-step Explanation

For the particle to move undeflected, the net Lorentz force must be zero: F = q (E + v \times B) = 0 E + v \times B = 0 ⟹ E = - (v \times B) = B \times v Taking the cross product with B on both sides: E \times B = (B \times v) \times B Using vector triple product a \times (b \times c) = b (a \cdot c) - c (a \cdot b) : E \times B = v (B \cdot B) - B (v \cdot B) Since v ⊥ B, the dot product v \cdot B = 0 . E \times B = v B^{2} v = \frac{E \times B}{B ^{2}}

Try it yourself before checking the explanation above.

A charged particle with charge q enters a region of constant, uniform and mutually orthogonal fields E and B with a velocity v perpendicular to both E and B, and comes out without any change in magnitude or direction of v . Then:

v = \frac{E \times B}{B ^{2}}

v = \frac{B \times E}{B ^{2}}

v = \frac{E \times B}{E ^{2}}

v = \frac{B \times E}{E ^{2}}

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