Question

What is the magnetic field strength required to make a proton with a speed of 5.0 x 10^(5) m/s follow a circular path of radius 0.0200 m?

Answer 1

Answer:

Magnetic field strength required for this is 0.25 T

Explanation:

As we know that the proton moves in circular path in uniform magnetic field

so the radius of the path of the circle is given as

$R = \frac{mv}{qB}$

here we know that

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

$m = 1.6 \times 10^{-27} kg$

$R = 0.0200 m$

$v = 5 \times 10^5 m/s$

now we have

$B = \frac{1.6 \times 10^{-27} (5 \times 10^5)}{(1.6 \times 10^{-19})(0.02)}$

so we have

$B = 0.25 T$