Question:medium

A proton moving with a velocity of $8\times 10^5$ 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 (Charge of proton = $1.6\times 10^{-19}$ C and mass of the proton = $1.66\times 10^{-27}$ kg)

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For a charged particle moving perpendicular to a magnetic field, the radius of the circular path is given by $R = mv/qB$. It is crucial to perform calculations using SI units (meters, kilograms, Coulombs, Teslas) and only convert to millitesla (mT) at the very end.

Updated On: Mar 30, 2026

500 mT
100 mT
200 mT
400 mT

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The Correct Option is B

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