
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
161.1k+ 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.
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