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tnbstudent
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Homework Statement
A wire loop with area 8.00 cm2 is placed inside a 20.0 cm long solenoid with 5.00×10^4 turns that carries a current of 1.10 A. The loop is concentric with the solenoid. If the length of the solenoid is stretched so that its length increases at a rate of 8.000e-2 m/s, with the number of turns remaining constant, what is the initial induced potential difference in the wire loop?
Homework Equations
Magnetic Field for Solenoid B = μ*I*n (n is number of loops per unit length)
E= ΦB/dt
Area circle = ∏r^2
The Attempt at a Solution
I know that the area of the circle is 0.0008m^2 and that is not changing
The angle is 90°
For E= ΦB/dt there are three things that can change to cause the change in B.
1- change in area so you have (dA)/dt * B* cosθ
2 - change in flux so you have A*(dB/dt)*cosθ
3 - change in the angle A*B*-sinθ
Since the question is asking about the flux through the wire loop and the change is occurring in the length of the solenoid - I do not think Area (A) is changing. I also don't think that the angle is changing. Since, the length of the solenoid is changing I think that magnetic field is what is changing as a function of time.
The formula for magnetic field is B = μ*I*n (n is number of loops per unit length). This is where I start getting messed up (at least I think this is where my troubles start)...
dB/dt=(4*∏X10^-7)*(1.1A)*[(5.0*10^4turns)/(.20m*8.0*10^-2m/s)
Another thought I had was to find B initial - which I found to be (0.346T) and then multiple it by 1/8.0*10^-2m/s so get dB/dt.
Any advice is appreciated!
Thanks