Question

At a particular instant, a proton at the origin has velocity < 5e4, -2e4, 0> m/s. You need to calculate the magnetic field at location < 0.03, 0.05, 0 > m, due to the moving proton. What is the vector r?

Answer 1

Answer:

$9.7\times 10^{-5} T$

Explanation:

Velocity =$5\times 10^4i-2\times 10^4j$

r=$0.03i+0.05j$

r=$\mid r\mid=\sqrt{(0.03)^2+(0.05)^2}=0.058$

v=$\mid V\mid=\sqrt{(5\times 10^4)^2+(-2\times 10^{4})^2}=5.39\times 10^{2}$

We know that

$B=\frac{mv}{qr}$

Where q=$1.6\times 10^{-19} C$

Mass of proton=$1.67\times 10^{-27} kg$

Using the formula

$B=\frac{1.67\times 10^{-27}\times 5.39\times 10^2}{1.6\times 10^{-19}\times 0.058}$

$B=9.7\times 10^{-5} T$