
A particle of mass $M$ and charge $Q$ moving with velocity $\vec{v}$ describes a circular path of the radius $R$ when subjected to a uniform transverse magnetic field of induction $B$. The work done by the field when the particle completes one full circle is
A.$BQv2\pi R$
B.$\left( \dfrac{M{{v}^{2}}}{R} \right)2\pi R$
C. Zero
D.$BQ2\pi R$
Answer
216.3k+ views
Hint: When a particle of mass $M$ and charge $Q$ moving with velocity $\vec{v}$ describes a circular path in a magnetic field, the magnetic force acting on the particle will be perpendicular to the magnetic field and its velocity. Therefore the kinetic energy of the particle remains constant.
Formula used:
Power is given by the scalar dot product of force and velocity vector.
$P=\vec{F}.\vec{v}$
$P=$power
$F=$magnetic force
$v=$velocity of particle
And work done,$w=P\times t$
$t=$time
Complete answer:
A particle experiences a magnetic force when moving through a magnetic field. When a charged particle of mass $M$ moves with velocity$\vec{v}$, perpendicular to the uniform magnetic field $B$, this particle follows the curved path until it forms a complete circular path.

Fig. $q$ charged particle completes a full circle.
Or, in other words, the magnetic force is always perpendicular to the velocity,$\vec{v}$so that the net power transferred by it to the particle will be zero.
$P=\vec{F}.\vec{v}$
Or, $P=Fv\cos \theta $
Or, $P=Fv\cos {{90}^{o}}$ [Since $\theta =$ the angle between the magnetic force and velocity]
Or, $P=0$
Again power $p$is defined as the rate at which work is done.
$P=\dfrac{work}{time}=\dfrac{W}{t}$
Or,$W=P\times t$
As we know from above the net power on the particle is zero, thus putting this value we get,
$W=0\times t=0$
Whenever the magnetic force is perpendicular to the velocity, it does not work on the charged particle. Then the kinetic energy and velocity remain constant. This is because the direction of motion is affected but not the magnitude of velocity.
Hence, the work done by the field when the particle completes one full circle is zero.
Thus, option (C) is correct.
Note:If an electron is in the region where the electric field, magnetic field, and direction of velocity are parallel to each other, then the force due to the magnetic field is zero. The charged particle will be accelerated by the electric field only.
Formula used:
Power is given by the scalar dot product of force and velocity vector.
$P=\vec{F}.\vec{v}$
$P=$power
$F=$magnetic force
$v=$velocity of particle
And work done,$w=P\times t$
$t=$time
Complete answer:
A particle experiences a magnetic force when moving through a magnetic field. When a charged particle of mass $M$ moves with velocity$\vec{v}$, perpendicular to the uniform magnetic field $B$, this particle follows the curved path until it forms a complete circular path.

Fig. $q$ charged particle completes a full circle.
Or, in other words, the magnetic force is always perpendicular to the velocity,$\vec{v}$so that the net power transferred by it to the particle will be zero.
$P=\vec{F}.\vec{v}$
Or, $P=Fv\cos \theta $
Or, $P=Fv\cos {{90}^{o}}$ [Since $\theta =$ the angle between the magnetic force and velocity]
Or, $P=0$
Again power $p$is defined as the rate at which work is done.
$P=\dfrac{work}{time}=\dfrac{W}{t}$
Or,$W=P\times t$
As we know from above the net power on the particle is zero, thus putting this value we get,
$W=0\times t=0$
Whenever the magnetic force is perpendicular to the velocity, it does not work on the charged particle. Then the kinetic energy and velocity remain constant. This is because the direction of motion is affected but not the magnitude of velocity.
Hence, the work done by the field when the particle completes one full circle is zero.
Thus, option (C) is correct.
Note:If an electron is in the region where the electric field, magnetic field, and direction of velocity are parallel to each other, then the force due to the magnetic field is zero. The charged particle will be accelerated by the electric field only.
Recently Updated Pages
Wheatstone Bridge Explained: Working, Formula & Uses

Young’s Double Slit Experiment Derivation Explained

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

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

Electricity and Magnetism Explained: Key Concepts & Applications

Chemical Properties of Hydrogen - Important Concepts for JEE 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

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Understanding Electromagnetic Waves and Their Importance

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Average and RMS Value in Electrical Circuits

