JEE MAIN - Physics (2020 - 6th September Morning Slot - No. 2)
A particle of charge q and mass m is moving with a velocity $$ - v\widehat i$$ (v $$ \ne $$ 0) towards a large screen
placed in the Y - Z plane at a distance d. If there is a magnetic field $$\overrightarrow B = {B_0}\widehat k$$
, the maximum value of v
for which the particle will not hit the screen is :
$${{2qd{B_0}} \over m}$$
$${{qd{B_0}} \over {3m}}$$
$${{qd{B_0}} \over {2m}}$$
$${{qd{B_0}} \over {m}}$$
Explanation
In uniform magnetic field particle moves in a circular path, if the radius of the circular path is 'd', particle will
not hit the screen.
r = $${{mv} \over {q{B_0}}}$$
To not collide, r < d
$$ \Rightarrow $$ $${{mv} \over {q{B_0}}}$$ < d
$$ \Rightarrow $$ v < $${{q{B_0}d} \over m}$$
$$ \therefore $$ vmax = $${{q{B_0}d} \over m}$$
r = $${{mv} \over {q{B_0}}}$$
To not collide, r < d
$$ \Rightarrow $$ $${{mv} \over {q{B_0}}}$$ < d
$$ \Rightarrow $$ v < $${{q{B_0}d} \over m}$$
$$ \therefore $$ vmax = $${{q{B_0}d} \over m}$$
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