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NEET 2012

Exam: NEET
Year: 2012
Subject: Physics
Chapter: Magnetic Effects of Current

NEET 2012 · Physics · Magnetic Effects of Current

An alternating electric field of frequency v , is applied across...

An alternating electric field of frequency

v

, is applied across the dees (radius

=R

) of a cyclotron that is being used to accelerate protons (mass

=m

). The operating magnetic field

(B)

used in the cyclotron and the kinetic energy

(K)

of the proton beam, produced by it, are given by

Options

A $B=\frac{m v}{e}$ and $K=2 m \pi^2 v^2 R^2$
B $B=\frac{2 \pi m v}{e}$ and $K=m^2 \pi v R^2$
C $B=\frac{2 \pi m v}{e}$ and $K=2 m \pi^2 v^2 R^2$
D $B=\frac{m v}{e}$ and $K=m^2 \pi v R^2$

Correct answer

C. $B=\frac{2 \pi m v}{e}$ and $K=2 m \pi^2 v^2 R^2$

Step-by-step solution

Frequency

v=\frac{e B}{2 \pi m}

\mathrm{KE}=\frac{1}{2} m v^2

and radius

R=\frac{m v}{e B}

Here, velocity

v=\frac{\pi R}{T / 2}=\frac{2 \pi R}{T}=2 \pi R v

\therefore \quad

Radius

R=\frac{m(2 \pi R v)}{e B}

Magnetic field

B=\frac{2 \pi m v}{e}

Kinetic energy

K=\frac{1}{2} m(2 \pi R v)^2=2 m \pi^2 v^2 R^2

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