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4.Moving Charges and Magnetism
medium
Electron of mass $m$ and charge $q$ is travelling with a speed along a circular path of radius $r$ at right angles to a uniform magnetic field of intensity $B$. If the speed of the electron is doubled and the magnetic field is halved the resulting path would have a radius
A
$2\,r$
B
$4\,r$
C
$\frac {r}{4}$
D
$\frac {r}{2}$
(AIIMS-2009)
Solution
Radius of path is given by $r=\frac{m v}{B q}$
Here, $\mathrm{m}$ and $\mathrm{q}$ remain unchanged
So, $\frac{r_{1}}{r_{2}}=\frac{v_{1}}{v_{2}} \cdot \frac{B_{2}}{B_{1}}=\frac{v}{2 v} \cdot \frac{B / 2}{B}=\frac{1}{4}$
$\Rightarrow r_{2}=4 r$
Standard 12
Physics