The two-loop wire circuit is 122.444 cm wide and 81.6292 cm high. The wire circu

Question

The two-loop wire circuit is 122.444 cm wide and 81.6292 cm high. The wire circuit in the figure is located in a magnetic field whose magnitude varies with time according to the expression B = (0.001 T/s) t and its direction is into the page.
Assume The resistance per length of the wire is 0.11 /m.

The two-loop wire circuit is 122.444 cm wide and 81.6292 cm high. The wire circuit in the figure is located in a magnetic field whose magnitude varies with time according to the expression B = (0.001 T/s) t and its direction is into the page. Assume The resistance per length of the wire is 0.11 ?/m. When the magnetic field is 0.6 T, find the magnitude of the current through middle leg P Q of the circuit. Answer in units of muA

Explanation / Answer

This looks trick, but let's start with the left loop. It has an area of:

0.816292m * 0.408146m = 0.333166m^2

So we can derive an expression for magnetic flux through this loop:

= BA

(t) = (0.333166m^2)(0.001)t

The emf in the loop is then

= d/dt

= (0.333166m^2)(0.001) = 3.33166*10^-4V

You can do a similar calculation for the emf in the right loop. It has an area of 0.666333m^2. Therefore its emf is 6.66333*10^-4V

The left loop contains 2.448876m of wire so its resistance is (0.11ohm/m)(2.448876) = 0.26937636ohms

So by ohms law the current in the left loop is

I = V/R = 3.33166*10^-4V / 0.26937636 = 0.0012368A

The right loop contains 3.265168m of wire so its resistance is 0.35916848ohms

A similar calculation reveals that its current is 0.0018552A

These currents are in opposite directions in the wire in question (via right hand rule) so the total current is:

(0.0018552A) - (0.0012368A) = 6.1841*10^-4A

OR

0.61841 mA

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