Find the magnitude of the induced current in the wire.

+3 votes

A piece of insulated wire is shaped into a figure eight as shown in the figure below. For simplicity, model the two halves of the figure eight as circles. The radius of the upper circle is 4.00 cm and that of the lower circle is 8.00 cm. The wire has a uniform resistance per unit length of 4.00 Ω/m. A uniform magnetic field is applied perpendicular to the plane of the two circles, in the direction shown. The magnetic field is increasing at a constant rate of 1.50 T/s.

image
(a) Find the magnitude of the induced current in the wire.
 A

(b) Find the direction of the induced current in the wire. (Select all that apply.)
 

 

asked Feb 28, 2012 in Physics by Joey33 ~Expert~ (1,216 points)
    

1 Answer

+4 votes
 
Best answer

From the prompt we know that:

radius of upper circle = 4.00 cm = 0.04 m

radius of lower circle = 8.00 cm = 0.08 m

λ = 4.00 Ω/m

DB/dt = 1.50 T/s


We can eventually solve this using Ohm's Law, but first we should find the values of V and R.

Magnetic Flux = A * B = (pi*(r^2 of upper loop)-pi*(r^2 of lower loop)*B

From Ampere's Law:

E = (dΦB)/(dt) = (pi*(r^2 of upper loop)-pi*(r^2 of lower loop) DB/dt

 

The resistance for the whole figure is:

R = λ (2 *pi * r of lower loop + 2 * pi * r of upper loop)

Now, with Ohm's Law (E = V = IR) we can plug in our vales and move them around to solve for I (I = V/R).

I = (V/R) = ((pi*(r^2 of upper loop)-pi*(r^2 of lower loop) DB/dt)/ λ (2 *pi * r of lower loop + 2 * pi * r of upper loop)

After simplifying:

I = (V/R) = (((r^2 of upper loop)-(r^2 of lower loop) DB/dt)/ (λ * 2 (r of lower loop + r of upper loop)))

 

Plug in your values and you should have the answer in units Amps.  From this the direction of the current should also be counterclockwise in the bottom, and so then also clockwise in the top since it has made a figure 8 shape.

answered Mar 3, 2012 by Hyperanthony ~Rookie~ (257 points)
selected Mar 3, 2012 by Joey33
Thanks great explanation!

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