JEE Main 2015 · 10 Apr · Shift 1 · Physics Q19

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PhysicsSingle correct+4 / -1Question 19 of 90

A proton (mass m) accelerated by a potential difference V flies through a uniform transverse magnetic field B. The field occupies a region of space by width 'd'. If 'α' be the angle of deviation of proton from initial direction of motion (see figure), the value of sin α will be :

A hatched strip of width d with field B into the page; the proton enters horizontally from the left and leaves on the right deflected upward by angle α.

\dfrac B2\sqrt{\dfrac{qd}{mV}}

\dfrac Bd\sqrt{\dfrac q{2mV}}

Bd\sqrt{\dfrac q{2mV}}

qV\sqrt{\dfrac{Bd}{2m}}

Answer and worked solution

Official answer: C

$v=\sqrt{\frac{2qV}m}$ and $r=\frac{mv}{qB}$ . Crossing a width $d<r$ , $\sin\alpha=\dfrac dr$ :
$\sin\alpha=\frac{dqB}{\sqrt{2qmV}}=Bd\sqrt{\frac q{2mV}}.$

Topics:Motion in magnetic and electric fields

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JEE Main 2015 · 10 Apr · Shift 1, Physics question 19