Verified Solution • 2025–26 Session moderate

QJEEM20Phfeef: JEE Main 2020 Physics - Class 12 JEEM20Phfeef Solution: An electron is moving along +x direction with a velocity of 6…

An electron is moving along +x direction with a velocity of 6 ×\times 106 ms–1. It enters a region of uniform electric field of 300 V/cm pointing along +y direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the x direction will be :

Multiple Choice Options

  • A 3 ×\times 10–4 T, along –z direction
  • B 5 ×\times 10–3 T, along –z direction
  • C 5 ×\times 10–3 T, along +z direction
  • D 3 ×\times 10–4 T, along +z direction

Step-by-Step Verified Solution

JEE Main 2020 (Online) 6th September Morning Slot Physics - Magnetic Effect of Current Question 127 English Explanation

B→\overrightarrow B must be in +z axis.

V→=6×106i^\overrightarrow V = 6 \times {10^6}\widehat i

E→=300j^\overrightarrow E = 300\widehat j V/cm = 3 ×\times 104 j^\widehat j V/m

F→\overrightarrow F = qE→+qV→×B→q\overrightarrow E + q\overrightarrow V \times \overrightarrow B

∴\therefore qE→+qV→×B→q\overrightarrow E + q\overrightarrow V \times \overrightarrow B = 0

⇒\Rightarrow qE = qVB

⇒\Rightarrow B = EV{E \over V} = 3×1046×106{{3 \times {{10}^4}} \over {6 \times {{10}^6}}} = 5 ×\times 10–3 T

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