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Physics Question on Moving charges and magnetism

A charged particle (electron or proton) is introduced at the origin (? = 0, ? = 0, ? = 0) with a given initial velocity v⃗\vec{v}. A uniform electric field E⃗\vec{E} and a uniform magnetic field B⃗\vec{B} exist everywhere. The velocity v⃗\vec{v}, electric field E⃗\vec{E} and magnetic field B⃗\vec{B} are given in columns 1, 2 and 3, respectively. The quantities ?0,?0?_0 , ?_0 are positive in magnitude. Column 1Column 2Column 3
(I) Electron with v⃗=2E0B0x^\vec{v}=2 \frac{E_{0}}{B_{0}}\hat{x}(i) E⃗=−E0z^\vec{E}=-E_{0}\hat{z}(P) B⃗=−B0x^\vec{B}=-B_{0}\hat{x}
(II)Electron with v⃗=E0B0y^\vec{v}= \frac{E_{0}}{B_{0}}\hat{y}(ii) E⃗=−E0y^\vec{E}=-E_{0}\hat{y}(Q) B⃗=−B0x^\vec{B}=-B_{0}\hat{x}
(III) Proton with v⃗=0\vec{v}=0(iii) E⃗=−E0x^\vec{E}=-E_{0}\hat{x}(R) B⃗=−B0y^\vec{B}=-B_{0}\hat{y}
(IV)Proton with v⃗=2E0B0x^\vec{v}=2 \frac{E_{0}}{B_{0}}\hat{x}(iv) E⃗=−E0x^\vec{E}=-E_{0}\hat{x}(S) B⃗=−B0z^\vec{B}=-B_{0}\hat{z}
In which case will the particle move in a straight line with constant velocity?
A

(III) (ii) (R)

B

(IV) (i) (S)

C

(III) (iii) (P)

D

(II) (iii) (S)

Answer

(II) (iii) (S)

Explanation

Solution

F⃗=q[E⃗+v⃗×B⃗]=−e[−E0x^+(E0B0y^)×(B0z^)]\vec{ F } = q [\vec{ E }+\vec{ v } \times \vec{ B }]=- e \left[- E _{0} \hat{ x }+\left(\frac{ E _{0}}{ B _{0}} \hat{ y }\right) \times\left( B _{0} \hat{ z }\right)\right]
=−e[−E0x^+E0x^]=0=- e \left[- E _{0} \hat{ x }+ E _{0} \hat{ x }\right]=0
∴\therefore Particle moves along a straight line ( yy-axis).