Emf of a single loop wire around a solenoid

AI Thread Summary
The discussion centers on calculating the magnetic flux through a wire loop surrounding a solenoid with specific dimensions and current. The solenoid has a diameter of 4 cm, a length of 20 cm, 250 turns, and carries a current of 15 A, while the wire loop has a diameter of 10 cm and is positioned perpendicular to the solenoid's axis. The magnetic field (B) generated by the solenoid is calculated using the formula B = (4π x 10^-7) * (number of turns) * (current), resulting in a value of 0.0264 T. The magnetic flux is then determined using the equation flux = B * A * cos(theta), yielding a final answer of 2.0735 x 10^-4 T*m^2. The confusion regarding the presence of an EMF is clarified, emphasizing that the question specifically asks for magnetic flux.
imatreyu
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Sorry, the title is wrong. Should be "magnetic flux" of a single loop wire around a solenoid.

Homework Statement



A solenoid 4 cm in diameter and 20 cm in length has 250 turns and carries a current of 15 A. It is surrounded by a loop of wire 10 cm in diameter. The loop is positioned perpendicular to and centered on the axis of the solenoid. Find the magnetic flux through the loop of wire. (The field outside of the solenoid is small enough to be negligible.)

Homework Equations



E= N*A*B*w*sinwt ?

The Attempt at a Solution



I don't know where to start...at all. How is there even an emf in the loop if the field around the solenoid is negligible?
 
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There's no EMF, but the question is asking for the flux. What's the definition of flux?
 
flux is BAcostheta.I'm confused on what to use for B. . .since there's no field created by the solenoid.
 
Are you sure the solenoid doesn't create a magnetic field?
 
Actually. . . if someone could check my work:

B= 4pi x 10^-7 * 1400 turns * 15 A
= .0264 T

flux = BA cos theta
=.0264 T * (.0025 pi m^2)
=2.0735 x 10^-4 T*m^2

Answer = 2.0735 x 10^-4 T*m^2
 
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