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JEE Advanced · Physics · 20. Magnetism & Current

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. A 2,iii(S)
  2. B 4,i(S)
  3. C 3,ii(R)
  4. D 3,iiiP
Verified Solution

Answer & Solution

Correct Answer

(A) 2,iii(S)

Step-by-step Solution

Detailed explanation

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
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