
A magnetic field
A. Always exerts a force on a charged particle
B. Never exerts a force on a charged particle
C. Exerts a force, if the charged particle is moving across the magnetic lines of force
D. Exerts a force, if the charged particle is moving along the magnetic lines of force
Answer
218.1k+ views
Hint: In the given question, we need to determine whether the magnetic field exerts a force or not in certain conditions. For this, first we will define the magnetic field. 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.
Magnetic force experienced by a charged particle is \[\vec F = q(\vec v \times \vec B)\].
Here, \[F\] is the force, \[B\] is the magnetic field, \[q\] is the charge of a particle and \[v\] is the velocity.
Complete answer:
We know that the magnetic force experienced by a charged particle is \[\vec F = q(\vec v \times \vec B)\].
Here, \[F\] is the force, \[B\] is the magnetic field, \[q\] is the charge of a particle and \[v\] is the velocity.
So, consider the following figure for this.

Image: Direction of cross product of magnetic field and velocity
We can define it as \[F = qvB\sin \theta \]
Here, we can say that If \[{\bf{v}}\parallel {\bf{B}}\] then only \[{\bf{F}} = {\bf{0}}\].
Thus, the velocity is parallel to the magnetic field line means the value of \[\sin \theta \] is zero.
This indicates that \[\theta = {0^o}\].
As a result, the magnitude of force is non-zero only if a charged particle moves across the magnetic field lines of forces.
Therefore, the correct option is (C).
Note: Many students forget to write an expression for the magnetic force experienced by a charged particle. They may give the explanation without the formula but it will not give a clear idea about the condition. This is the only way through which we can solve the example in the simplest way.
Formula used:
The following formula is used for solving the given question.
Magnetic force experienced by a charged particle is \[\vec F = q(\vec v \times \vec B)\].
Here, \[F\] is the force, \[B\] is the magnetic field, \[q\] is the charge of a particle and \[v\] is the velocity.
Complete answer:
We know that the magnetic force experienced by a charged particle is \[\vec F = q(\vec v \times \vec B)\].
Here, \[F\] is the force, \[B\] is the magnetic field, \[q\] is the charge of a particle and \[v\] is the velocity.
So, consider the following figure for this.

Image: Direction of cross product of magnetic field and velocity
We can define it as \[F = qvB\sin \theta \]
Here, we can say that If \[{\bf{v}}\parallel {\bf{B}}\] then only \[{\bf{F}} = {\bf{0}}\].
Thus, the velocity is parallel to the magnetic field line means the value of \[\sin \theta \] is zero.
This indicates that \[\theta = {0^o}\].
As a result, the magnitude of force is non-zero only if a charged particle moves across the magnetic field lines of forces.
Therefore, the correct option is (C).
Note: Many students forget to write an expression for the magnetic force experienced by a charged particle. They may give the explanation without the formula but it will not give a clear idea about the condition. This is the only way through which we can solve the example in the simplest way.
Recently Updated Pages
Young’s Double Slit Experiment Derivation Explained

Wheatstone Bridge Explained: Working, Formula & Uses

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE General Topics in Chemistry Important Concepts and Tips

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

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

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

Understanding Atomic Structure for Beginners

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

How to Convert a Galvanometer into an Ammeter or Voltmeter

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Understanding Electromagnetic Waves and Their Importance

Understanding Average and RMS Value in Electrical Circuits

