29.38     A long, thin solenoid has 900 turns per meter and radius 2.50 cm.  The current in the solenoid is increasing at a uniform rate of 36.0 A/s.  What is the magnitude of the induced electric field at a point near the center of the solenoid and(a) 0.500 cm from the axis of the solenoid;(b) 1.00 cm from the axis of the solenoid?

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Asked Nov 8, 2019
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29.38     A long, thin solenoid has 900 turns per meter and radius 2.50 cm.  The current in the solenoid is increasing at a uniform rate of 36.0 A/s.  What is the magnitude of the induced electric field at a point near the center of the solenoid and

(a) 0.500 cm from the axis of the solenoid;

(b) 1.00 cm from the axis of the solenoid?

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Step 1

Part A:

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Given Info: The number of turns (n) per meter of a long thin solenoid is 900. The radius of the solenoid coil is 2.50 cm. The current in the solenoid increasing at a uniform rate (dI/dt) is 36.0 A/s. The induced electric field at a point near the center of the solenoid with distance r from its axis is 0.500 cm. Sketch the solenoid end view as follows, r 0.500 cm Figure

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Step 2
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Formula to calculate the induced electric field by Maxwell's equation is, dp Ed dt dB is the rate of change in magnetic flux, is the differential length of circumference of the solenoid. Take magnitude on both sides, E.dl dt rdl in above equation, Substitute 2Tr for dB E (2nr) dt Rearrange the above equation as follows, dФв 1 E= 2лт (I) dt

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Step 3
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Formula to calculate the magnitude of the induced emf is, dB Eind (II) dt dB is the rate of change in magnetic flux with respect to time. dt аФр. Substitute Eind| for in equation (I) to find E. dt E Eind 2лт Rearrange the above equation, lEind = 2TrE (II) Formula to calculate the magnetic flux is Фв — ВА Bis the uniform magnetic field, A is the constant area of the solenoid. in equation (II) to find leind| Substitute BA for |d(BA) Eind dt (IV) - A E dt

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