
If a proton is projected in a direction perpendicular to a uniform magnetic field with velocity \[v\] and an electron is projected along the magnetic field, what will happen to proton and electron?
A. The electron will travel along a circle with constant speed and also the proton will move along a straight line.
B. A proton will move in a circular path with constant speed and there will be no effect on the motion of the electron.
C. There will not be any side effects on the motion of the electron and proton.
D. The electron and proton both will follow the path of a parabola.
Answer
162k+ views
Hint: In the given question, we need to find the condition of the proton and electron. For this, we need to use the formula for force experienced by a charged particle in an external magnetic field to get the desired result.
Formula used:
The following formula is used for solving the given question.
The magnetic force on moving charge is given by
\[\vec F = q(\vec v \times \vec B) = qvBsin\theta \]
Here, \[F\] is the force, \[v\] is the velocity, \[B\] is the magnetic field strength, and \[q\] is charge.
Complete answer:
We know that the magnetic force on the moving charge is \[\vec F = q(\vec v \times \vec B) = qvBsin\theta \]
Here, \[F\]is the force, \[v\] is the velocity, \[B\] is the magnetic field strength, and \[q\] is charge.
So, for Proton, we can say that \[\theta = {90^ \circ }\]
Also, for electron, \[\theta = {0^ \circ }\]
Therefore, maximum force will act on the proton in a direction perpendicular to both \[\vec v\] and \[\vec B\], so it will move along a circular path.
Thus, we can say that force on an electron will be zero because it is moving parallel to the field.
Therefore, the correct option is (B).
Note: Many students make mistakes in writing the formula of the magnetic force on the moving charge. This is the only way through which we can solve the example in the simplest way. Also, it is essential to analyze the formula carefully to get the desired result regarding proton and electron.
Formula used:
The following formula is used for solving the given question.
The magnetic force on moving charge is given by
\[\vec F = q(\vec v \times \vec B) = qvBsin\theta \]
Here, \[F\] is the force, \[v\] is the velocity, \[B\] is the magnetic field strength, and \[q\] is charge.
Complete answer:
We know that the magnetic force on the moving charge is \[\vec F = q(\vec v \times \vec B) = qvBsin\theta \]
Here, \[F\]is the force, \[v\] is the velocity, \[B\] is the magnetic field strength, and \[q\] is charge.
So, for Proton, we can say that \[\theta = {90^ \circ }\]
Also, for electron, \[\theta = {0^ \circ }\]
Therefore, maximum force will act on the proton in a direction perpendicular to both \[\vec v\] and \[\vec B\], so it will move along a circular path.
Thus, we can say that force on an electron will be zero because it is moving parallel to the field.
Therefore, the correct option is (B).
Note: Many students make mistakes in writing the formula of the magnetic force on the moving charge. This is the only way through which we can solve the example in the simplest way. Also, it is essential to analyze the formula carefully to get the desired result regarding proton and electron.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

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

Young's Double Slit Experiment Step by Step Derivation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Charging and Discharging of Capacitor

Wheatstone Bridge for JEE Main Physics 2025

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
