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An electron of mass $m$ and charge $q$ is travelling with a speed $v$ along a circular path of radius $r$ at right angles to a uniform magnetic field $B$. If speed of the electron is doubled and the magnetic field is halved, then resulting path would have a radius of
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The correct answer is:
$4 r$
In a perpendicular magnetic field, Magnetic force $=$ centripetal force ie,
$\begin{aligned}
& B q v=\frac{m v^2}{r} \Rightarrow r=\frac{m v}{B q} \\
& \therefore \quad \frac{r_1}{r_2}=\frac{v_1}{v_2} \times \frac{B_2}{B_1} \\
& \frac{r_1}{r_2}=\frac{1}{2} \times \frac{1}{2} \\
& r_2=4 r_1 \\
& \Rightarrow \quad r_2=4 r \\
&
\end{aligned}$
$\begin{aligned}
& B q v=\frac{m v^2}{r} \Rightarrow r=\frac{m v}{B q} \\
& \therefore \quad \frac{r_1}{r_2}=\frac{v_1}{v_2} \times \frac{B_2}{B_1} \\
& \frac{r_1}{r_2}=\frac{1}{2} \times \frac{1}{2} \\
& r_2=4 r_1 \\
& \Rightarrow \quad r_2=4 r \\
&
\end{aligned}$
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