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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 π2ml5qE
B time taken by particle to cross region -I for 2nd time is π2ml13qE
C Angular speed of 1st revolution in magnetic field is 5qE2ml
Using work-energy theorem till string become taut
Work done by qE = change in K.E
qEl=12mv20
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=12mv212m(vcos30)2
qEl2=12mv212m3qEl2m
qEl2=12mv2123qEl2
mv2=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=π2ml5qE
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=mv212m25qEl2m
v1=13qEl2m
time period to cross region I for second time = πlv1=π2ml13qE
Angular speed of 1st half revolution in region I= vl=5qE2ml
Angular speed of 2nd half revolution in region = v1l=13qE2ml

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