A proton accelerated by a potential difference $V=500\,\text{kV}$ moves through a transverse magnetic field $B=0.51\,\text{T}$ as shown in the figure. The width of the magnetic field region is $d=10\,\text{cm}$. Then the angle $\theta$ through which the proton deviates from the initial direction of its motion is (approximately)
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A charged particle entering a uniform magnetic field perpendicular to its velocity moves in a circular path of radius $r=\dfrac{mv}{qB}$.
Step 1: Find the speed of the proton after acceleration through potential $V$.
\[
\frac{1}{2}mv^2 = qV
\Rightarrow v=\sqrt{\frac{2qV}{m}}
\]
For a proton:
\[
q=1.6\times10^{-19}\,\text{C},\quad
m=1.67\times10^{-27}\,\text{kg},\quad
V=5\times10^5\,\text{V}
\]
\[
v=\sqrt{\frac{2(1.6\times10^{-19})(5\times10^5)}{1.67\times10^{-27}}}
\approx 9.8\times10^6\,\text{m s}^{-1}
\]
Step 2: Radius of circular path in magnetic field:
\[
r=\frac{mv}{qB}
=\frac{1.67\times10^{-27}\times 9.8\times10^6}{1.6\times10^{-19}\times0.51}
\approx 0.20\,\text{m}
\]
Step 3: The proton travels an arc of a circle inside the field.
From geometry (see figure),
\[
\sin\theta=\frac{d}{r}
\]
Here,
\[
d=0.10\,\text{m},\quad r\approx0.20\,\text{m}
\]
\[
\sin\theta=\frac{0.10}{0.20}=0.5
\Rightarrow \theta=30^\circ
\]
