A charged particle 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

Question

A charged particle 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 a uniform magnetic field $\vec{B}$ are given in the columns below

Column 1	Column 2	Column 3
1. Electron with $\vec{v} = 2 \frac{E_{0}}{B_{0}} \hat{x}$	(i) $\vec{E} = E_{0} \hat{z}$	(P) $\vec{B} = - B_{0} \hat{x}$
2. Electron with $\vec{v} = \frac{E_{0}}{B_{0}} \hat{y}$	(ii) $\vec{E} = - E_{0} \hat{y}$	(Q) $\vec{B} = B_{0} \hat{x}$
3.Proton with $\vec{v}$ =0	(iii) $\vec{E} = {- E}_{0} \hat{x}$	(R) $\vec{B} = - B_{0} \hat{y}$
4. Proton with $\vec{v} = 2 \frac{E_{0}}{B_{0}} \hat{x}$	(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 a constant velocity?

TestHub · Accepted Answer

Given: A charged particle is introduced at the origin ( x = 0 , y = 0 , z = 0 ) . initial velocity = v , uniform electric field E . uniform magnetic field B . Also the values are given the table. we need to find the cases where the particle will move in a straight line with constant velocity. To move a particle with constant velocity; Fnet = 0 , ⇒ The electric force = magnetic fore e. qE = q v B . E = v B . ​ Now checking the options. ​ A) (II) (iii) (s). (II) = v = B 0 ​ E 0 ​ ​ y ^ ​ . (iii) = E = − E 0 ​ x ^ ( s ) = B = B 0 ​ Z ^ . − E 0 ​ x ^ = E 0 ​ BO ^ y ^ ​ ( BO z ^ ) ​ ∴ (II) (iii) (s) is the correct option

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