Question

A proton enters a uniform magnetic field of intensity 10T with a velocity $3\times {10}^{7}m/s$ in a direction at an angle of ${90}^{\circ }$ with the magnetic field. The radius of the circle traced by the proton is

• A. 3.1 cm
• B. 30 cm
• C. 1 cm
• D. 100 cm

Answer 1

The correct option is A 3.1 cm

Change on the proton, $q=1.6\times {10}^{-19}C$
Magnetic field, B = 10 T
Velocity of the proton, $v=3\times {10}^{7}m/s$
Angle between magnetic field and velocity of proton, $\theta ={90}^{\circ }$
Mass of the proton, $m=1.67\times {10}^{-27}kg$
Now, centripetal force on the proton = magnetic force on the proton
$\Rightarrow \frac{m{v}^2}{r}=qvBsin\theta =qvB(\because sin{90}^{\circ }=1)$
$\Rightarrow r=\frac{mv}{qB}=\frac{1.67\times {10}^{-27}\times 3\times {10}^7}{1.6\times {10}^{-19}\times 10}$
$=3.1\times {10}^{-2}m$
= 3.1 cm
Hence, the correct answer is option (1).