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## Homework Statement

A square, single-turn wire coil L = 1.75 cm on a side is placed inside a solenoid that has a circular cross section of radius r = 2.82 cm.

http://capa8.phy.ohiou.edu/res/ohiou/serwaylib/Graphics/Graph20/serw2014.gif

The solenoid is 22.0 cm long and wound with 113 turns of wire. If the current in the solenoid is 3.29 A, find the flux through the coil.

## Homework Equations

Є=(ΔNΔΦ/Δt)

ΔΦ=BAcosΘ

B=μ0NI

## The Attempt at a Solution

I have found the strength of the magnetic field using the constant μ0 = 12.57x10^-7, the number of turns 113, and the current 3.29 A, and because time is not applied Δt can be taken out of the equation to get Є=ΔNΔΦ=ΔN(BAcosΘ). This gives me B = 4.6731489x10^-4 or B = 4.67x10^-4 The next part of the question asks for the flux and the strength of the magnetic field is required to do so. However, when applying the strength of the magnetic field and the area of the square coil using the equation Φ = BA the answer I am given is Φ = 1.43x10^-7 Wb which is apparently wrong. I'm not sure what im doing wrong and any help with this would be greatly appreciated.

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