In the picture at the right is shown a solenoid and two conducting loops. When t

Question

In the picture at the right is shown a

solenoid and two conducting loops.

When the switch is closed, the solenoid

carries a current in the direction

indicated by the orange arrows. The

planes of the conducting loops are

parallel to the planes of the loops of the

solenoid.

Conducting loop #1 consists of a single

turn of resistive wire that has a

resistance per unit length of ?.

Conducting loop #2 consists of N turns

of the same wire. Each loop is a circle

of radius r

a. The switch is closed and

remains closed for a few

seconds. Conducting loop #1 is

then moved to the right. Is

there a current flow induced?

If there is a current, indicate

the direction and explain how

you figured it out.

b. Conducting loop #1 is now returned to its original location and both conducting loops are held

fixed. The switch is opened. For a short time, the magnetic field felt uniformly at both conducting

loops decreases like

B = B0 - ?t

where ? is a constant. Is there a current flow induced in the loops? If there is a current, indicate

the direction

c. Calculate the current flow in each loop for situation (b) in terms of ?, ?, r, B0, ?, and t.

d. If B0 = 10 G, ? = 2.5 G/s, the resistivity of the wire is 1 microOhm/m, and the loops have a radius

of 2 cm, calculate the current induced in loop #1 as the B field from the solenoid begins to fall.

In the picture at the right is shown a solenoid and two conducting loops. When the switch is closed, the solenoid carries a current in the direction indicated by the orange arrows. The planes of the conducting loops are parallel to the planes of the loops of the

Explanation / Answer

1. When the switch is closed the field of the blue solenoid obeys the right screw rule (it has the direction of a right screw rotated as the current turns). The induction B field vector is from left to right.

a) when one moves the small loop the field produced by the big solenoid decreases with distance. Thus there will be a variation of the total magnetic flux, through the small loop surface. As Farady law states there will be an induced current, that creates a field opposing the decreasing of the initial field. Since the field created by the induced current has the same direction, the current creating it has the same direction as the current in the big blue solenoid (it is rotating counterclockwise).

b) The total flux in any of the small loops will change because the field change.

d(Fi) = d(B)*S

U = -d(Fi)/dt = -(d(B)/dt) *S = gamma*S =gamma*pi*r^2

(for loop #2 this is the induced volatge per 1 turn).

Again the same rule applies: the current in the loops tend to compensate for the decreasing of the field produced by the big blue solenoid. The induced currents will create a filed from left to right, the current will rotate counterclockwise (the same as in the big blue coil -rule of right screw).

c) because loop #1 has only one turn

U =gamma*pi*r^2 =2.5*10^-4*pi*0.02^2 =3.14*10^-7 V

(1 Gauss =10^-4 Tesla)

because loop #2 has N turns

U = N*gamma*pi*r^2

The resistance of loop #1 is

R1 = 2*pi*r*rho = 2*pi*0.02*1*10^-6 =1.26*10^-7 Ohm

The current is

I = U/R = 2.49 A

The resistance of loop #2 is

R2 =N*R1

The current is the same as above.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.