Time remaining:
What is the induced current in the loop?

label Physics
account_circle Unassigned
schedule 0 Hours
account_balance_wallet $5

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

Did you know? You can earn $20 for every friend you invite to Studypool!
Click here to
Refer a Friend
...
Apr 17th, 2015
...
Apr 17th, 2015
Sep 26th, 2017
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer