Question

A proton of mass $1.67\times 10^{-27}kg$ enters a uniform magnetic field $1T$ of at point A shown in figure with a speed of $10^{7}m{s}^{-1}$
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. $0.7m,45^{\circ }$
• B. $0.7m,90^{\circ }$
• C. $0.14m,90^{\circ }$
• D. $0.14m,45^{\circ }$

Answer 1

Answer: (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
$AC-2r\cos 45^{\circ }-2r\times \frac{1}{\sqrt{2}}-\sqrt{2}r$ ...(i)
As $Bqv=\frac{m{v}^2}{r}$ or $r=\frac{mv}{Bq}$
$AC=\frac{\sqrt{2}mv}{Bq}=\frac{\sqrt{2}\times 1.67\times 10^{-27}\times 10^7}{1\times 1.6\times 10^{-19}}$
$=0.14m$