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What is the induced current in the loop?

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A 19cm length of 0.27mm diameter nichrome wire is welded into a circular loop. The loop is placed between the poles of an electromagnet, and a field of 0.55T is switched on in a time of 14ms


Apr 17th, 2015

The current in the circuit is equal to the induced electromotive force E.M.F. divided by the resistance of the loop.

I = Ɛi/R

The electromotive force is calculated by Faraday's law:

Ɛi = ΔфB/∆t 

where  фB Is the magnetic flux through the loop. The magnetic flux is the total number of field lines, through the area occupied by the loop and has the following value: 

фB = B A = B πr^2. 

Here, πr^2 is the area of the loop of radius r. The radius of the loop can be computed from the total length of the wire:

2πr = L  →  r = L/2π   

r^2 = L^2/4π^2

Evaluating the equation of magnetic flux, we have:

фB = (B π)(L^2/4π^2) = BL^2/4π

Initially, the magnetic field in the elecroiman is zero. After closing the switch during 14ms, the EMF generated is:

Ɛi = ΔфB/∆t  = (BL^2/4π)/t = BL^2/4πt

Ɛi = (0.55)(0.19)^2/4(3.14)(0.014) = 0.113 V

To find the resistance R, we apply the law of Pouillet:

R = ρL/S

Where ρ = 10^-6 Ω m, is the resistivity of nichrome, L is the length of wire and S is the area of cross section.

The wire radius is:  r = d/2 = (2.7 10^-4)/2 = 1.35 10^-4 m

R = ρL/πr^2 = (10^-6)(0.19)/(3.14)(1.35 10^-4)^2

R = 3.32 Ω

Finally, the current generated is:

I = Ɛi/R = 0.113/3.32 = 0.034 A = 34 mA

Apr 18th, 2015

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