A 100pF capacitor is connected to a coil with tow loops. In this drawing the loo

Question

A 100pF capacitor is connected to a coil with tow loops. In this drawing the loops are perpendicular to the page. The plates of the capacitor are separated by a gap of 1.0mm. A magnetic field pointing to the right is increased in magnitude. Which plate of the capacitor (left/right) acquires a positive charge? Dry air is known to break down and allow electrical sparking at an electric field of about 30E6 V/m. What is the minimum rate of change in the magnetic flux that will cause sparking across the capacitor?

Explanation / Answer

the flux in coil increases

the magnetic field is right

induced emf decreases the flux

induced magnetic field is left

current is entering to left plate

so left acquires + charge

LEFT

B)

potential difference V = Ed

V = Einduced = rate of change in flux

rate of change in flux = E*d = 3*10^6*10^-3 = 3000 Wb/s

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