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JEE Advanced 2017Paper 1
Exam
JEE Advanced
Year
2017
Paper
Paper 1
Subject
Physics
Chapter
Magnetic Effects of Current

JEE Advanced 2017 · Paper 1 · Physics · Magnetic Effects of Current

A charged particle is introduced at the origin x = 0, y = 0, z = 0...

A charged particle is introduced at the origin x=0,y=0,z=0x = 0, y = 0, z = 0 with a given initial velocity v→\overset{\to}{v} . A uniform electric field E→\overset{\to}{E} and a uniform magnetic field B→\overset{\to}{B} exist everywhere. The velocity v→\overset{\to}{v} , electric field E→\overset{\to}{E} and a uniform magnetic field B→\overset{\to}{B} are given in the columns below
Column 1 Column 2 Column 3
1. Electron with \overset{\to}{v} = 2\dfrac{{E}_{0}}{{B}_{0}}\overset{^}{x} (i) \overset{\to}{E} = {E}_{0}\overset{^}{z} (P) \overset{\to}{B} = - {B}_{0}\overset{^}{x}
2. Electron with \overset{\to}{v} = \dfrac{{E}_{0}}{{B}_{0}}\overset{^}{y} (ii) \overset{\to}{E} = - {E}_{0}\overset{^}{y} (Q) \overset{\to}{B} = {B}_{0}\overset{^}{x}
3.Proton with v→\overset{\to}{v} =0 (iii) \overset{\to}{E} = {- E}_{0}\overset{^}{x} (R) \overset{\to}{B} = - {B}_{0}\overset{^}{y}
4. Proton with \overset{\to}{v} = 2\dfrac{{E}_{0}}{{B}_{0}}\overset{^}{x} (iv) \overset{\to}{E} = {E}_{0}\overset{^}{x} (S) \overset{\to}{B} = - {B}_{0}\overset{^}{z}
In which case will the particle move in a straight line with a constant velocity?

Options

  1. A(2),(iii)(S)\left(2\right), \left(iii\right) ( S )
  2. B(4),(i)(S)\left(4\right), \left(i\right) ( S )
  3. C(3),(ii)(R)\left(3\right), \left(ii\right) ( R )
  4. D(3),(iii)(P)\left(3\right), \left(iii\right)\left(P\right)

Correct answer

A. (2),(iii)(S)\left(2\right), \left(iii\right) ( S )

Step-by-step solution

Given: A charged particle is introduced at the origin (x=0,y=0,z=0)(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=0, (\Rightarrow) The electric force == magnetic fore e. (\begin{gathered} q E=q v B . \\ E=v B . \end{gathered}) Now checking the options.  A) (II) (iii) (s).  (II) =v⃗=E0B0y^.  (iii) =E⃗=−E0x^(s)=B⃗=B0Z^. −E0x^=E0BO^y^(BOz^)\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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