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A charged particle having kinetics energy K enters into the region of a uniform magnetic field between two plates P and Q as shown in fig. The charged particle just misses hitting the plate Q. The magnetic field in the region between the two plates??

Syed suha , 7 Years ago
Grade 12
anser 2 Answers
Sandeep Bhamoo

Last Activity: 7 Years ago

For such movement to occur radius of curvature of its motion should be equal to the distance "d" r=d=mv/qB B= mv/qd = √2mK/qd that is fourth option. Please do approve my answer if correct

Prince Raj

Last Activity: 3 Years ago

We know..
K=½mv²
From here;find v,i.e
V=√(2k/m)..............(1)
 
Now...
We know that, magnetic force
Fm=qBV...........(2)
Here Fm is equal to centripetal force(due to circular motion of charge particle).
So,
MV²/r=qBV
MV/r=qB
 
Put the value of V from eqn (1)
[M*√(2k/m)]/r=qB
B={M√(2k/m)}/qr
B=[√(2km)] / qd  Ans.
 

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