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A charge of 2 μC revolves with a frequency of 6.25 × 10¹² Hz and creates a magnetic field of 6.28 T at a point on the circle. What is the radius of the circular path?

2.25 m
0.25 m
13.0 cm
1.25 m

Correct answer: 1.25 m

Solution

I = q*f = 2e-6 * 6.25e12 = 1.25e7 A. From B = mu0*I/(2r), r = mu0*I/(2B) = (4*pi*1e-7 * 1.25e7)/(2*6.28) = 1.25 m.

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