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
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
B
sin1(23)
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
C
30o
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
D
180o
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

## The correct option is D 180oLet magnetic field strength be BCharge qMass m and Radius of the circular path be r→B=B^k→v=vcosθ^i+vsinθ^j→F=(vcosθ^i+vsinθ^j)×Bq^k⇒→F=Bqv(cosθ^i+sinθ^j)θ=ωtax=Bqvmsinωtd2xdt2=BqvmsinωtIntegrating⇒x=Bqvmω2sin(ωt+θ)So magnitude of maximum horizontal displacement suffered by the particle in the field will be xmax=Bqvmω2Again,xmax=Bqvmω2Bqv=mω2rω=Bqmxmax=vωv=rω⇒xmax=rThis means for a particle to reach maximum displacement the field must have a minimum length of rsince 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°

Suggest Corrections
1
Join BYJU'S Learning Program
Related Videos
Magnetic Field Due to a Current in a solenoid
PHYSICS
Watch in App
Join BYJU'S Learning Program