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BITSAT 2015 Paper

Section: Physics

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Question : 7 of 150

Marks: +1, -0

A charged particle moving in a uniform magnetic field and losses 4% of its kinetic energy. The radius of curvature of its path changes by

2%
4%
10%
12%

Solution:

As we know F = qvB =

\frac{mv^{2}}{r}

∴ r =

\frac{mv}{Bq}

And KE = k =

\frac{1}{2} mv^{2}

∴ mv =

\sqrt{2km}

∴ r =

\frac{mv}{qB}

=

\frac{\sqrt{2km}}{qB}

⇒ r α

\sqrt{k}

or r =

c^{1/2}

(c is a constant)

\frac{dr}{r}

= c

\frac{dk^{1/2}}{dr}

or

\frac{c\Delta k}{\Delta r}

= 2

\sqrt{k}

or

\frac{\Delta r}{r}

=

\frac{c\Delta k}{2\sqrt{k}c\sqrt{k}}

=

\frac{\Delta k}{2k}

Therefore percentage changes in radius ofpath,

\frac{\Delta r}{r}

× 100 =

\frac{\Delta k}{2k}

× 100 = 2%

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