Question

A proton travels through a region of uniform magnetic field at an angle \thetaθ relative to the magnetic field. The magnitude of the magnetic field in this region is 1.17 T, and the proton's velocity is 500,000 m/s when it experiences an acceleration whose magnitude is 1.50 \times 10^{13}~\frac{m}{s^2}1.50×10 ​13 ​​ ​s ​2 ​​ ​ ​m ​​ . Calculate the angle \thetaθ

Answer 1

Answer:

$\theta=15.52^{\circ}$

Explanation:

The magnitude of the force of a moving cahrge, in our case a proton, trought a magentic field is given by:

$F=|q||vB|sin(\theta)$ (1)

where:

q is the proton charge ($q=1.6*10^{-19} C$)

v is the proton velocity ($v=5*10^{5} m/s$)

B is the magnetic field (B = 1.17 T)

Now, we just need to solve the equaton (1) for \theta.

$\theta=sin^{-1}\left(\frac{F}{qvB}\right)$

But the force F = ma, then:

m is the mass of proton  ($m=1.67*10^{-27} kg$)

a is the acceleration ($a=1.5*10^{13} m/s^{2}$)

$\theta=sin^{-1}\left(\frac{m_{p}a}{q_{p}vB}\right)$

$\theta=sin^{-1}\left(\frac{1.67*10^{-27}*1.5*10^{13}}{1.6*10^{-19}*5*10^{5}*1.17}\right)$

$\theta=15.52^{\circ}$

I hope it helps you!