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

B=0.15T

Step-by-step 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=(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 = ((1.67*10^(-27)kg)(3*10^(6)(m)/(s)))/((1.6*10^(-19)C)(0.2m))\\\\B=0.15T`

Hope this helps!!

Answer 2

0.16T

Step-by-step explanation: