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Question

# Consider a bob of mass m and having charge q attached with light string of length l and pivoted at point O. It is released at rest at 60∘ with vertical. There are two regions- Region- I (left of line PQ) has a uniform and constant magnetic field B directed into plane of paper. Region-II (right of line PQ) has a constant and uniform electric field E directed vertically up as shown. Consider no effect of gravity in both the regions.

A
time taken by particle to cross region -I for 1st time is π2ml5qE
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B
time taken by particle to cross region -I for 2nd time is π2ml13qE
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C
Angular speed of 1st revolution in magnetic field is 5qE2ml
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D
Angular speed of 2nd revolution in magnetic field is 13qEml
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Solution

## The correct options are A time taken by particle to cross region -I for 1st time is π√2ml5qE B time taken by particle to cross region -I for 2nd time is π√2ml13qE C Angular speed of 1st revolution in magnetic field is √5qE2mlUsing work-energy theorem till string become taut Work done by qE = change in K.E ⇒qEl=12mv2−0 ⇒v=√2qElm This speed v is vertically upwards. When string becomes taut an impulse is received and speed (tangential) of the bob becomes vcos30∘=√3qEl2m Further using work energy theorem Work done by qE = Change in K.E, qEl2=12mv′2−12m(vcos30∘)2 qEl2=12mv′2−12m3qEl2m qEl2=12mv′2−123qEl2 mv′2=5qEl2 speed of the bob as it enters magnetic field v′=√5qEl2m For the first half revolution in the magnetic field particle will move with same constant speed v′ ∴ time period to cross region I for first time =πlv′=π√2ml5qE After 1st half revolution in magnetic field, particle enters the region of electric field. Due to work done by electric field there is change in KE. Let v1 be velocity of particle as it enters region-I for the second time. Work done by qE = Change in K.E, from lowermost to uppermost point. qE.2l=mv212−m25qEl2m v1=√13qEl2m time period to cross region I for second time = πlv1=π√2ml13qE Angular speed of 1st half revolution in region I= v′l=√5qE2ml Angular speed of 2nd half revolution in region = v1l=√13qE2ml

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