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A long, thin solenoid has 870 turns per meter and radius 3.00cm . The current in...

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Physics

A long, thin solenoid has 870 turns per meter and radius 3.00cm. The current in the solenoid is increasing at a uniform rate of65.0A/s .

A)What is the magnitude of the induced electric field at a point0.550cm from the axis of the solenoid?

B)What is the magnitude of theinduced electric field at a point 1.00cm from the axis of thesolenoid?

Answer & Explanation Solved by verified expert
4.0 Ratings (618 Votes)

a.

The magnetic field due to solenoid is,

           B = uonI

Differentiate the above equation on both sides,

          dB/dt = uon dI/dt

The expression for the electric field inside the solenoid is,

       E = r/2 dB/dt

           = (r/2)uon dI/dt

           = (0.550x10-2 m /2)(4(pi)x10-7 T.m/A)(870)(65.0 A/s)

           = 1.95x10-4 N/C

-----------------------------------------------------------------------------------

b.

The expression for the electric field inside the solenoid is,

       E = r/2 dB/dt

           = (r/2)uon dI/dt

           = (1.00x10-2 m /2)(4(pi)x10-7 T.m/A)(870)(65.0 A/s)

           = 3.55x10-4 N/C
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