A piece of insulated wire is shaped into a figure eight as shown in the figure b

Question

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 7.00 cm. The wire has a uniform resistance per unit length of 5.00 /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.

(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.)

clockwise in the upper loopclockwise

in the lower loopcounterclockwise

in the upper loopcounterclockwise

in the lower loop

Explanation / Answer

The Magnetic Field B at any time t is given by :

B(t) = B0 + kt, where k = 1.5T/s

Lets define the two cirlces as Cu and Cl .

Choosing the area vector A of both the circles to be pointing into the page, we have Magnetic flux of each circle at any time t as:

u = B(t) Au= B(t)* (4cm)2

l = B(t) Al= B(t)* (7cm)2

Using Faraday's Law, Induced emf in each of the circles is given by :

u= - du/dt = - k * (4cm)2

l= - dl/dt = - k * (7cm)2

Since both the loops are continuous, net induced emf will be the modulus of the difference of the two emf's.

Hence, net induced Emf, = - k * 33

Induced Current, I = I I / R

R = (Total Length of the Insulated wire) * 5/m = 2* * ( 0.04m+0.07m) *5/m

R = (1.1*)

Induced Current, I = I I / R = k * 33 / (1.1*)

I = 1.5 * 30 = 45A

Hence, the net induced current in the loop is 45A.

The direction is anticlockwise in the lower circle and clockwise in the upper circle.

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