Question

A proton having a speed of 3.0 × 106 m/s in a direction perpendicular to a uniform magnetic field moves in a circle of radius 0.20 m within the field. What is the magnitude of the magnetic field?

Answer 1

Answer:

0.16T

Explanation:

Answer 2

Answer:

B=0.15T

Explanation:

In a magnetic field we have that the force generated by the magnetic field equals the centripetal force

Fm=qvB=mv^2/r

Hence, this problem can be solved using the expression that relates the radius of a curved path described by a particle and a magnetic field

$B=\frac{m_pv}{q_pr}$

mp: mass of a proton = 1.67*10^{-27}kg

qp: charge of the proton = 1.6*10^{-19}C

r: radius = 0.20m

v: speed of the proton = 3*10^{6}m/s

- By replacing in the expression we have

$B = \frac{(1.67*10^{-27}kg)(3*10^{6}\frac{m}{s})}{(1.6*10^{-19}C)(0.2m)}\\\\B=0.15T$

Hope this helps!!