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

(A) $\vec{v} = \dfrac{\vec{E}\times\vec{B}}{B^{2}}$

(B) $\vec{v} = \dfrac{\vec{B}\times\vec{E}}{B^{2}}$

(C) $\vec{v} = \dfrac{\vec{E}\times\vec{B}}{E^{2}}$

(D) $\vec{v} = \dfrac{\vec{B}\times\vec{E}}{E^{2}}$

JEE Main · 2007 · 29 Apr 2007 · Offline · Magnetic Effects of Current

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

(A) $v = \frac{E \times B}{B ^{2}}$ ✓ Correct
(B) $v = \frac{B \times E}{B ^{2}}$
(C) $v = \frac{E \times B}{E ^{2}}$
(D) $v = \frac{B \times E}{E ^{2}}$

Answer: (A)

Solution

For the particle to pass through undeflected the electric and magnetic forces must cancel, so

q E + q v \times B = 0

, that is

E = - v \times B

. Taking the cross product of this with

B

and using that

v

is perpendicular to

B

gives

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

. The selected velocity has magnitude

\frac{E}{B}

and the direction set by

E \times B

JEE Main · 2007 · 29 Apr 2007 · Offline · Single correct option

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