A charged particle (electron or proton) is introduced at the origin $(x=0, y=0, z=0)$ with a given initial velocity $\vec{v}$. A uniform electric field $\vec{E}$ and a uniform magnetic field $\vec{B}$ exist everywhere. The velocity $\vec{v}$, electric field $\vec{E}$ and magnetic field $\vec{B}$ are given in columns 1, 2 and 3, respectively. The quantities $E_{0}, B_{0}$ are positive in magnitude. | Column 1 | Column 2 | Column 3 | | --- | --- | --- | | (I) Electron with $\vec{v}=2 \frac{E_{0}}{B_{0}} \widehat{x}$ | (i) $\vec{E}=E_{0} \hat{z}$ | (P) $\vec{B}=-B_{0} \hat{x}$ | | (II) Electron with $\vec{v}=\frac{E_{0}}{B_{0}} \hat{y}$ | (ii) $\vec{E}=-E_{0} \hat{y}$ | (Q) $\vec{B}=B_{0} \hat{x}$ | | (III) Proton with $\vec{v}=0$ | (iii) $\vec{E}=-E_{0} \hat{x}$ | (R) $\vec{B}=B_{0} \hat{y}$ | | (IV) Proton with $\vec{v}=2 \frac{E_{0}}{B_{0}} \widehat{x}$ | (iv) $\vec{E}=E_{0} \hat{x}$ | (S) $\vec{B}=B_{0} \hat{z}$ | In which case will the particle move in a straight line with constant velocity?

JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Magnetic Effects of Current

A charged particle (electron or proton) is introduced at the origin

(x = 0, y = 0, z = 0)

with a given initial velocity

v

. A uniform electric field

E

and a uniform magnetic field

B

exist everywhere. The velocity

v

, electric field

E

and magnetic field

B

are given in columns 1, 2 and 3, respectively. The quantities

E_{0}, B_{0}

are positive in magnitude.

Column 1	Column 2	Column 3
(I) Electron with $v = 2 \frac{E _{0}}{B _{0}} x$	(i) $E = E_{0} \overset{z}{^}$	(P) $B = - B_{0} \overset{x}{^}$
(II) Electron with $v = \frac{E _{0}}{B _{0}} \overset{y}{^}$	(ii) $E = - E_{0} \overset{y}{^}$	(Q) $B = B_{0} \overset{x}{^}$
(III) Proton with $v = 0$	(iii) $E = - E_{0} \overset{x}{^}$	(R) $B = B_{0} \overset{y}{^}$
(IV) Proton with $v = 2 \frac{E _{0}}{B _{0}} x$	(iv) $E = E_{0} \overset{x}{^}$	(S) $B = B_{0} \overset{z}{^}$

In which case will the particle move in a straight line with constant velocity?

Answer: (D)

For a charged particle to move with constant velocity, the net force must be zero:

q [E + (v \times B)] = 0

. For option (D): (II) (iii) (S) corresponds to an electron with

v = \frac{E _{0}}{B _{0}} \overset{y}{^}

E = - E_{0} \overset{x}{^}

, and

B = B_{0} \overset{z}{^}

. The magnetic force is

F_{B} = - e (v \times B) = - e (\frac{E _{0}}{B _{0}} \overset{y}{^} \times B_{0} \overset{z}{^}) = - e E_{0} \overset{x}{^}

. The electric force is

F_{E} = - e E = - e (- E_{0} \overset{x}{^}) = e E_{0} \overset{x}{^}

. The net force is

F_{E} + F_{B} = e E_{0} \overset{x}{^} - e E_{0} \overset{x}{^} = 0

, so the particle moves with constant velocity.

JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Single correct option