Question

A long black coil has N1=1600 turns over its length of l=80.0 cm, and the radius of r= 4.0 cm. The current I running through this coil flows in CCW direction. The blue coil is wrapped around the black one, and it has the total of 5 full turns. At the ends of the blue coil is a small resistor with the resistance of 0.1 Ohms.

A.Find the net flux through the blue coil. Give your answer in terms of quantities: N1, l, r, I, and N2. Show all work

B. If the current in the black coil begins to increase steadily, so that over the 2.0 sec period the current goes up from 1.0 A to 1.4 A, calculate the induced EMF in the blue coil, and the induced current. Tell the direction of the induced current and explain why it is so.

Answer 1

A.

The number of turns per unit length of the long black coil is

n = N₁ / L

   = 1600 turns / (80.0) ( 1 m / 100 cm)

    =2000 turns /m

The magnetic field produced by the long black coil is

B =₀nI

Substitute 1.26 x 10-6 m kg/s²A² for mo and 2000 turns /m for n in the above equation,

It is given that the blue coil is 5 turns. The total surface area linked with the blue coil is’

A = 5 (r²)

Thus, the net flux linked with the blue coil is

Substitute 4.0 cm for r in the above equation,

Therefore, the net flux linked with the blue coil is (6.33 x 10^-6I) Wb.

2.

The change in the current is

I = 1.4 A – 1.0 A

   = 0.4 A

Time interval is

t = 2.0 s

Now, the induced emf is equal to the number of turns of the blue coil times the rate of change of current in the black coil.

Thus, the induced emf is

V = (5 turns) ( 0.4 A / 2.0 s)

    = 4.0 V

Now, the induce current in the blue coil is

I_i = V/R

   = 4.0 V / 0.1 ohms

= 40 A

Therefore, the induced emf in the blue coil is 4.0 V and current is 40A.

The direction of the induced current in the blue coil will be opposite to the current flow in the black coil.