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BITSAT 2015

Exam: BITSAT
Year: 2015
Subject: Physics
Chapter: Magnetic Effects of Current

BITSAT 2015 · Physics · Magnetic Effects of Current

A charged particle moving in a uniform magnetic field and losses 4...

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

Options

A $2 \%$
B $4 \%$
C $10 \%$
D $12 \%$

Correct answer

A. $2 \%$

Step-by-step solution

As we know

\mathrm{F}=\mathrm{qvB}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}\therefore \mathrm{r}=\frac{\mathrm{mv}}{\mathrm{Bq}}

And

\mathrm{KE}=\mathrm{k}=\frac{1}{2} \mathrm{mv}^{2}\therefore \mathrm{mv}=\sqrt{2 \mathrm{ km}}\therefore \mathrm{r}=\frac{\mathrm{mv}}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{ km}}}{\mathrm{qB}}\Rightarrow r \propto \sqrt{k}

or

r=c^{1 / 2}(c

is a constant

)\frac{\mathrm{dr}}{\mathrm{dr}}=\mathrm{c} \frac{\mathrm{dk}^{1 / 2}}{\mathrm{dr}}

or

\frac{\mathrm{c} \Delta \mathrm{k}}{\Delta \mathrm{r}}=2 \sqrt{\mathrm{k}}\frac{\Delta r}{r}=\frac{\mathrm{c} \Delta \mathrm{k}}{2 \sqrt{\mathrm{k}} \mathrm{c} \sqrt{\mathrm{k}}}=\frac{\Delta \mathrm{k}}{2 \mathrm{k}}

Therefore percentage changes in radius of path,

\frac{\Delta r}{r} \times 100=\frac{\Delta k}{2 k} \times 100=2 \%

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