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This is solution to problems related Electrical Engineering course. It was suggested by Prof. Bhooshan Sawhney at Shree Ram Swarup College of Engineering and Management. It includes: Radius, Capacitor, Distance, Axis, Points, Magnitude, Magnetic, Field, Maximum, Result
Typology: Exercises
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B = μ^0 ε^0 r 2
dE dt
and for r ≥ R, it is B = μ 0 ε 0 R^2 2 r
dE dt
The maximum magnetic field occurs at points for which r = R, and its value is given by either of the formulas above: Bmax = μ 0 ε 0 R 2
dE dt
There are two values of r for which B = Bmax/2:one less than R and one greater. To find the one that is less than R, we solve μ 0 ε 0 r 2
dE dt
μ 0 ε 0 R 4
dE dt for r. The result is r = R/2 = (55.0 mm)/2 = 27.5 mm. To find the one that is greater than R, we solve
μ 0 ε 0 R^2 2 r
dE dt =^
μ 0 ε 0 R 4
dE dt for r. The result is r = 2R = 2(55.0 mm) = 110 mm.