Question

At what speed (in m/s) will a proton move in a circular path of the same radius as an electron that travels at 8.00 ✕ 106 m/s perpendicular to the Earth's magnetic field at an altitude where the field strength is 1.20 ✕ 10−5 T?

Answer 1

Answer:

4347.8 m/s  

Explanation:

It is given that the radius of the circular path traversed by proton and electron is same. Also, we know that magnitude of charge on an electron and proton is same. Magnetic field strength is same for both.

$\frac{m_ev_e^2}{r}=qv_eB\\\frac{m_pv_p^2}{r}=qv_pB$

Take the ratio:

$m_ev_e=m_pv_p\\\Rightarrow v_p=\frac{m_e}{m_p}v_e\\\Rightarrow v_p=\frac{1}{1840}\times 8.0\times 10^6 m/s\\\Rightarrow v_p=4347.8m/s$