Answer to Question #101728 in Electricity and Magnetism for Sridhar

Question

Two particles carrying equal charges move parallel to each other with speed 150 km/s. If F(1) and F(2) are magnetic and electric forces between two charged particles then the ratio of F(1)/ F(2) is
Ans 25×10^-8

Answer 1

$q_1=q_2$

$v_1=v_2=v=150\times 10^3 \; m/s$

$c^2=8.98 \times 10^{16} \; m^2/s^2$

$F_1$ magnetic force

$F_2$ electric force

$\frac{F_1}{F_2} -?$

Solution:

Suppose that $q_1>0, q_2>0$ ( if $\;q_1<0, q_2<0$ solution will be similar)

Electric force: $\;\; F_2=k\frac{q_1 q_2 }{r^2}, \quad k=\frac{1}{4\pi \varepsilon _0}$

Magnetic force:$\;\;F_1=q_2 v_2 B_1\sin\theta, \quad \theta =90^{\circ}$ angle between $v_2$ and $B_1$

$\;\;F_1=q_2v B_1=q_2v\frac{\mu _0}{4\pi}\frac{q_1v}{r^2}=\frac{\mu _0}{4\pi}\frac{q_1 q_2 v^2}{r^2}$

$\frac{F_1}{F_2}=\mu_0 \varepsilon_0 v^2 =\lbrack \; c^2=\frac{1}{\varepsilon_0 \mu_0} \; \rbrack= \frac{v^2}{c^2}$

$\frac{F_1}{F_2}= \frac{(150\times 10^3)^2}{8.98\times 10^{16}}=25\times 10^{-8}$

Answer: $25\times 10^{-8}.$