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Question

A particle of charge q and mass m moving under the influence of uniform electric field E ^i and a uniform magnetic field −B ^i moves from point P (0,a) to Q (2a,a) where O (0,0) is origin. The velocity of the particle at Point P is v ^i and at Q is 2v ^i. The rate of work done by the electric field & magnetic field on particle at point P will be respectively:

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Solution

The correct option is **A** 34mv3a,0

The motion of particle will be a straight line because the velocity of particle is parallel to direction of →E and antiparallel to direction of →B. The magnetic force acting on particle will be,

→F=q(→v×→B)=0

Thus workdone by magnetic force will be zero throughout motion of charge.

Increase in kinetic energy of particle,

ΔKE=12m(2v)2−12m(v2)

ΔKE=32mv2

Workdone by →E in going from P to Q is,

We=Fe×d

We=qE×(2a)=2qEa

Using Work-energy theorem,

We+Wm=ΔKE

or, 2qEa+0=32mv2

E=3mv24qa

Now the rate of work done is power delivered and it will be only due to electric field, because the power delivered by magnetic field is zero.

⇒Pe=→Fe.→v

Pe=(qE ^i).(v ^i)=qEV

Pe=q(3v2m4qa)v=3mv34a

Pm=0

The motion of particle will be a straight line because the velocity of particle is parallel to direction of →E and antiparallel to direction of →B. The magnetic force acting on particle will be,

→F=q(→v×→B)=0

Thus workdone by magnetic force will be zero throughout motion of charge.

Increase in kinetic energy of particle,

ΔKE=12m(2v)2−12m(v2)

ΔKE=32mv2

Workdone by →E in going from P to Q is,

We=Fe×d

We=qE×(2a)=2qEa

Using Work-energy theorem,

We+Wm=ΔKE

or, 2qEa+0=32mv2

E=3mv24qa

Now the rate of work done is power delivered and it will be only due to electric field, because the power delivered by magnetic field is zero.

⇒Pe=→Fe.→v

Pe=(qE ^i).(v ^i)=qEV

Pe=q(3v2m4qa)v=3mv34a

Pm=0

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