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  4. A proton (mass m) accelerated by a potential difference V flies through a uniform transverse magnetic field B. The field occupies a region of space by width d. If α be the angle of deviation of proton from initial direction of motion (see figure), the value of sin α will be :

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Q. A proton (mass m) accelerated by a potential difference $V$ flies through a uniform transverse magnetic field $B$. The field occupies a region of space by width $d$. If $\alpha$ be the angle of deviation of proton from initial direction of motion (see figure), the value of $\sin \,\alpha$ will be :Physics Question Image

JEE MainJEE Main 2015Moving Charges and Magnetism

Solution:

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$q V=\frac{1}{2} m v^{2}$
$\Rightarrow v=\sqrt{\frac{2 q v}{m}} r=\frac{m v}{q B}=\frac{m}{q B} \sqrt{\frac{2 q V}{m}}$
$r=\frac{1}{B} \sqrt{\frac{2 m V}{q}} \sin \alpha$
$=\frac{d}{r}=B d \sqrt{\frac{q}{2 m V}}$