A proton moving with a velocity of $8 \times 10^5 \mathrm{~ms}^{-1}$ enters a uniform magnetic field normal to the direction of the magnetic field. If the radius of the circular path of the proton in the magnetic field is 8.3 cm, then the magnitude of the magnetic field is $\left(\right.$ Charge of proton $=1.6 \times 10^{-19} \mathrm{C}$ and mass of the proton $\left.=1.66 \times 10^{-27} \mathrm{~kg}\right)$

• A

  500 mT
• Correct answer
  B

  100 mT
• C

  200 mT
• D

  400 mT

Hence, the correct answer is option: 'B'.