A proton moving in a circular path perpendicular to a constant magnetic field takes 1.00 µs tocomplete one revolution. Determine the magnitude of the magnetic field. Taken as q =1.602x10-19 C and m= 1.67x10-27 kg for proton and r=3.14.0.0345 TA)0.0327 TB)0.0143 T0.0124 TD)0.0154 TE)

Time taken (T) = 1 μsq = 1.602×10-19 C mass (m) = 1.67×10-27 kg

Answered: A proton moving in a circular path…

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A proton moving in a circular path perpendicular to a constant magnetic field takes 1.00 µµs to complete one revolution. Determine the magnitude of the magnetic field. Taken as q = 1.602x10-19 C and m= 1.67x1027 kg for proton and t=3.14. 0.0345 T A) 0.0327 T B) 0.0143 T C) 0.0124 T D) 0.0154 T