A proton which moves perpendicular to a magnetic field of 1.2tin a circular path

Question

A proton which moves perpendicular to a magnetic field of 1.2tin a circular path of.08m radius, has what speed? a. 3.4 E(6) m/s b. 4.6 E(6) m/s c. 9.6 E(6) m/s d. 9.2 E(6) m/s Can someone show me the steps to this one? I got c but do notthink I did it right. A proton which moves perpendicular to a magnetic field of 1.2tin a circular path of.08m radius, has what speed? a. 3.4 E(6) m/s b. 4.6 E(6) m/s c. 9.6 E(6) m/s d. 9.2 E(6) m/s Can someone show me the steps to this one? I got c but do notthink I did it right.

Explanation / Answer

The speed of the proton is       v = (Bqr)/m Here m =1.67*10-27kg            B = 1.2 T            r = 0.08 m            q =1.6*10-19 C So v = 9.2 E(6) m/s So option (d ) is correct.

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