Question

A uniform magnetic field $ B = 1.2 \,mT $ is directed vertically upward throughout the volume of a laboratory chamber. A proton $ (m_p =1.67 \times 10^{-27}\, kg) $ enters the laboratory horizontally from south to north. Calculate the magnitude of centripetal acceleration of the proton if its speed is $ 3 \times 10^7\, m/s $ .

• A. $ 3.45 \times 10^{12}\,m/s^{2} $
• B. $ 1.67 \times 10^{12}\,m/s^{2} $
• C. $ 5.25 \times 10^{12}\,m/s^{2} $
• D. $ 2.75 \times 10^{12}\,m/s^{2} $

Answer 1

Answer: (A)

Centripetal acceleration $a_{c}=\frac{F_{m}}{m_{p}}$
$=\frac{Bqv}{m_{p}}$
$=\frac{1.2\times10^{-3}\times1.6\times10^{-19}\times3\times10^{7}}{1.67\times10^{-27}}$
$=3.6\times10^{12} m/ s$