A long solenoid has n = 375 turns per meter and carries a current given by I = 2

Question

A long solenoid has n = 375 turns per meter and carries a current given by I = 29.5(1 e1.60t ), where I is in amperes and t is in seconds. Inside the solenoid and coaxial with it is a coil that has a radius of R = 6.00 cm and consists of a total of N= 250 turns of fine wire (see figure below). What emf is induced in the coil by the changing current?

Part 1 of 5 - Conceptualize

The solenoid and the flat coil are not part of the same circuit. The transfer of energy between them is mediated by the magnetic field created by the solenoid. The solenoid current starts from zero, increases most rapidly at first, and then asymptotically approaches 29.5 A as its final value. The induced voltage will start from its largest value, on the order of one volt, at time zero, and then decay exponentially to zero.

Part 2 of 5 - Categorize

The current in the solenoid will be used to find its field. We will use Faraday's law to calculate the induced emf.

Part 3 of 5 - Analyze

We are given the number of turns per meter and the exponential expression for the current over time. Remember that

0

is a constant, the permeability of free space. The solenoid creates a magnetic field given by

Part 4 of 5 - Analyze

The magnetic flux through one turn of the flat coil is B =

, but since

dA cos

refers to the area perpendicular to the flux, and the magnetic field is uniform over the area A of the flat coil, the integral simplifies to the following.

B = 0nI

= 4 × 107 N/A2
turns/m A 1 e1.60t

= × 102 N/m · A 1 e1.60t .

Explanation / Answer

As per guide lines I am working first problem

the magnetic field inside solenoid is

B = uo nI

the magnetic flux through the coil is

flux = BA = uo nI pi r^2

from the Faraday's law

E = - N d flux/dt = - N uo n pi r^2 d I/ dt = - N uo n pi r^2 d/dt ( Io e^-1.6 t)

=uo N n pi r^2 Io ( 29.5) e^-1.6 t

= 4 pi * 10^-7 ( 250)(375) pi ( 0.06)^2 ( 29.5) ( 1.6) e^-1.6 t

=0.0628253712e^-1.6 t

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