Q.
A proton of mass $1.67 \times 10^{- 27}$ kg enters a uniform magnetic field of 1 T at point $A$ as shown in figure, with a speed of $10^{7}\ ms^{- 1}$. The magnetic field is directed normal to the plane of paper downwards. The proton emerges out of the magnetic field at point $C$, then the distance $AC$ and the value of angle $\theta$ will respectively be
A) 0.7 m, $45{^\circ}$
B) 0.7 m, 90${^\circ}$
C) 0.14 m, 90${^\circ}$
D) 0.14 m, 45${^\circ}$
Explanation
(d) From the symmetry of figure, the angle $\theta = 45{^\circ}$. The path of moving proton in a normal magnetic field is circular. If $r$ is the radius of the circular path, then from the figure,
$AC = 2\ r\cos{45{^\circ} = 2r \times \frac{1}{\sqrt{2}} = \sqrt{2}r}\ $ …(i)
As $B\ q\ V = \frac{mv^{2}}{r}$ or $r = \frac{mv}{Bq}$
$AC = \frac{\sqrt{2}\ mv}{B\ q} = \frac{\sqrt{2} \times 1.67 \times 10^{- 27} \times 10^{7}}{1 \times 1.6 \times 10^{- 19}} = 0.14\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.