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NEET 2012

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

NEET 2012 · Physics · Magnetic Effects of Current

A proton carrying 1 MeV kinetic energy is moving in a circular path...

A proton carrying

1 \mathrm{MeV}

kinetic energy is moving in a circular path of radius

R

in uniform magnetic field. What should be the energy of an

\alpha

-particle to describe a circle of same radius in the same field?

Options

A $2 \mathrm{MeV}$
B $1 \mathrm{MeV}$
C $0.5 \mathrm{MeV}$
D $4 \mathrm{MeV}$

Correct answer

B. $1 \mathrm{MeV}$

Step-by-step solution

For proton

r=\frac{\sqrt{2 m(\mathrm{KE})}}{q B}

So

\quad q \propto \sqrt{m(\mathrm{KE})}

Hence

\begin{aligned}\frac{e}{2 e}& =\sqrt{\frac{\left(m_p\right)(1 \mathrm{MeV})}{\left(4 m_p\right)(\mathrm{KE})}}\\ \frac{1}{4}& =\frac{1 \mathrm{MeV}}{4 \mathrm{KE}}\\ \mathrm{KE}& =1 \mathrm{MeV} \end{aligned}

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