A charged particle is introduced at the origin x = 0 ,   y = 0 ,   z = 0 with a given initial…

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→ , electric field E→  and a uniform magnetic field B→ are given in the columns below
Column 1 Column 2 Column 3
1. Electron with  v→=2E0B0x^  (i) E→=E0z^ (P)  B→=-B0x^
2. Electron with  v→=E0B0y^ (ii)  E→=-E0y^ (Q)  B→=B0x^
3.Proton with v→ =0 (iii)  E→=-E0x^ (R)  B→=B0y^
4. Proton with  v→=2E0B0x^ (iv)  E→=E0x^ (S)  B→=B0z^
In which case will the particle move in a straight line with a constant velocity?
  1. 2,iii(S)
  2. 4,i(S)
  3. 3,ii(R)
  4. 3,iiiP

Solution

Given: A charged particle is introduced at the origin \((x=0, y=0, z=0)\). initial velocity \(=\vec{v}\), uniform electric field \(\vec{E}\). uniform magnetic field \(\vec{B}\). 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 \(=0\), \(\Rightarrow\) The electric force \(=\) magnetic fore e. \(\begin{gathered} q E=q v B . \\ E=v B . \end{gathered}\) Now checking the options. \(\begin{aligned} &\text { A) (II) (iii) (s). } \\ &\text { (II) }=\vec{v}=\frac{E_0}{B_0} \hat{y} \text {. } \\ &\text { (iii) }=\vec{E}=-E_0 \hat{x} \\ &(s)=\vec{B}=B_0 \hat{Z} \text {. } \\ &-E_0 \hat{x}=E_0 \hat{B O} \hat{y}(B O \hat{z}) \\ \end{aligned}\) \(\therefore\) (II) (iii) (s) is the correct option `

Asked in: JEE Advanced 2017 (Paper 1)

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