Question

A proton of mass $ 1.67 \times 10^{-27}\, kg$ enters a uniform magnetic field $1\, T$ of at point A 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. $0.14\, m ,\, 45^{\circ}$

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
$A C-2 r \cos 45^{\circ}-2 r \times \frac{1}{\sqrt{2}}-\sqrt{2} r$ ...(i)
As $B q v=\frac{m v^{2}}{r}$ or $ r=\frac{m v}{B q}$
$A C =\frac{\sqrt{2} m v}{B q}=\frac{\sqrt{2} \times 1.67 \times 10^{-27} \times 10^{7}}{1 \times 1.6 \times 10^{-19}}$
$=0.14\, m$