
An electron is moving along +x direction with a velocity of $6 × 10^6\,ms^{–1}$. It enters a region of the 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
A. \[3 \times {10^{ - 4}}{\rm{ T}}\], along –z direction
B. \[5 \times {10^{ - 3}}{\rm{ T}}\], along –z direction
C. \[5 \times {10^{ - 3}}{\rm{ T}}\], along +z direction
D. \[3 \times {10^{ - 4}}{\rm{ T}}\], along +z direction
Answer
163.5k+ views
Hint:Suppose any charge q is moving in the direction of the positive x- axis with velocity v. Because of charge moving an electric field exists around it. This electric field will apply a force on the charge whose magnitude will be qE in the direction that depends on the nature of the charge. If you assume the charge as positive. The electric field is trying to change the direction of the charge as it applies a force. To set up a magnetic field in the region to stop the direction of the charge. The net force must be in the opposite direction of the electric field.
Formula used
Force due to electric field,
\[{F_e} = q\overrightarrow E \]
Force due to magnetic field ,
\[{F_m} = q\overrightarrow {{\rm{ }}v} \times \overrightarrow B \]
In cross product, \[\widehat i \times \overrightarrow k = - \widehat j\]
Complete step by step solution:
Given velocity is along +x direction and the uniform electric field of 300 V/cm pointing along +y direction.

Image: Three dimensions of charge e moving
To calculate the direction of the magnetic field:
Force due to electric field is applied in the direction of \[\widehat j\],
\[{F_e} = q\overrightarrow E \]
Force due to the magnetic field must be in the opposite direction of the same magnitude as \[ - \widehat j\] ,
\[{F_m} = q\overrightarrow {{\rm{ }}v} \times \overrightarrow B \]
Thus
\[\begin{array}{l}\overrightarrow v \times \overrightarrow B = - \widehat j\\\widehat i \times \overrightarrow B = - \widehat j\\\widehat i \times \overrightarrow B = \widehat i \times \widehat k\\\overrightarrow B = \widehat k\end{array}\]
Therefore the direction of the magnetic field set up in this region will be along +z direction.
To calculate the direction of magnetic field:
Equating the magnitude as
\[{F_m} = {F_e}\]
\[qvB = qE\] (The charge of electron must be canceled)
Given velocity = $6 × 10^6\,ms^{–1}$
uniform electric field = 300 V/cm = \[3 \times {10^4}V/m\]
Substituting the value, we get
\[B = 5 \times {10^{ - 3}}T\]
Therefore the magnitude of the magnetic field set up in this region is \[B = 5 \times {10^{ - 3}}T\].
Hence option C is the correct answer.
Note: A Moving electron produces a magnetic field around it. Electric fields and magnetic fields are also known as electromagnetic fields or simply EMF. Which consist of waves of electric and magnetic energies moving together.
Formula used
Force due to electric field,
\[{F_e} = q\overrightarrow E \]
Force due to magnetic field ,
\[{F_m} = q\overrightarrow {{\rm{ }}v} \times \overrightarrow B \]
In cross product, \[\widehat i \times \overrightarrow k = - \widehat j\]
Complete step by step solution:
Given velocity is along +x direction and the uniform electric field of 300 V/cm pointing along +y direction.

Image: Three dimensions of charge e moving
To calculate the direction of the magnetic field:
Force due to electric field is applied in the direction of \[\widehat j\],
\[{F_e} = q\overrightarrow E \]
Force due to the magnetic field must be in the opposite direction of the same magnitude as \[ - \widehat j\] ,
\[{F_m} = q\overrightarrow {{\rm{ }}v} \times \overrightarrow B \]
Thus
\[\begin{array}{l}\overrightarrow v \times \overrightarrow B = - \widehat j\\\widehat i \times \overrightarrow B = - \widehat j\\\widehat i \times \overrightarrow B = \widehat i \times \widehat k\\\overrightarrow B = \widehat k\end{array}\]
Therefore the direction of the magnetic field set up in this region will be along +z direction.
To calculate the direction of magnetic field:
Equating the magnitude as
\[{F_m} = {F_e}\]
\[qvB = qE\] (The charge of electron must be canceled)
Given velocity = $6 × 10^6\,ms^{–1}$
uniform electric field = 300 V/cm = \[3 \times {10^4}V/m\]
Substituting the value, we get
\[B = 5 \times {10^{ - 3}}T\]
Therefore the magnitude of the magnetic field set up in this region is \[B = 5 \times {10^{ - 3}}T\].
Hence option C is the correct answer.
Note: A Moving electron produces a magnetic field around it. Electric fields and magnetic fields are also known as electromagnetic fields or simply EMF. Which consist of waves of electric and magnetic energies moving together.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Main 2025 Session 2: Exam Date, Admit Card, Syllabus, & More

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Trending doubts
Degree of Dissociation and Its Formula With Solved Example for JEE

Charging and Discharging of Capacitor

Instantaneous Velocity - Formula based Examples for JEE

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

In which of the following forms the energy is stored class 12 physics JEE_Main

JEE Main Chemistry Question Paper with Answer Keys and Solutions

Other Pages
Three mediums of refractive indices mu 1mu 0 and mu class 12 physics JEE_Main

Total MBBS Seats in India 2025: Government College Seat Matrix

NEET Total Marks 2025: Important Information and Key Updates

Neet Cut Off 2025 for MBBS in Tamilnadu: AIQ & State Quota Analysis

Karnataka NEET Cut off 2025 - Category Wise Cut Off Marks

NEET Marks vs Rank 2024|How to Calculate?
