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Question

A charged particle enters a magnetic field at right angles to the field. The field exists for a length equal to 1.5 times the radius of circular path of particle. The particle will be deviated from its path by :

A
90o
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B
sin1(23)
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C
30o
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D
180o
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Solution

The correct option is D 180o

Let magnetic field strength be B

Charge q

Mass m and

Radius of the circular path be r

B=B^k

v=vcosθ^i+vsinθ^jF=(vcosθ^i+vsinθ^j)×Bq^kF=Bqv(cosθ^i+sinθ^j)θ=ωtax=Bqvmsinωtd2xdt2=Bqvmsinωt

Integrating

x=Bqvmω2sin(ωt+θ)

So magnitude of maximum horizontal displacement suffered by the particle in the field will be

xmax=Bqvmω2

Again,

xmax=Bqvmω2Bqv=mω2rω=Bqmxmax=vω

v=rωxmax=r

This means for a particle to reach maximum displacement the field must have a minimum length of r

since length of field is 1.5r the particle cannot come out of field horizontally so to come out of it must reverse its direction and thus displacement.

Thus the particle will deflect by 180°


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