Answer:
The magnitude of the magnetic field is 0.156 T.
Explanation:
Given that,
Speed of a proton, [tex]v=3\times 10^6\ m/s[/tex]
Radius of the circular path, r = 0.2 m
When the proton enters in the circular path, the centripetal force is balanced by the magnetic force such that :
[tex]\dfrac{mv^2}{r}=qvB\sin\theta[/tex]
q is the charge on proton
Here, [tex]\theta=90^{\circ}[/tex]
[tex]\dfrac{mv^2}{r}=qvB\\\\B=\dfrac{mv}{qr}\\\\B=\dfrac{1.67\times 10^{-27}\times 3\times 10^6}{1.6\times 10^{-19}\times 0.2}\\\\B=0.156\ T[/tex]
So, the magnitude of the magnetic field is 0.156 T. Hence, this is the required solution.
