How Does Rewinding a Coil Affect Its Self-Inductance and Energy Storage?

[Soaring Crane]

Homework Statement

After you measure the self-inductance of a coil, you unwind it and then rewind half the length of wire into a coil with the same diameter but half the number of turns. How does this change the self-inductance?

a. It is the same.
b. It is doubled.
c. It is quadrupled.
d. It is halved.
e. It is quartered.

Homework Equations

See below.

The Attempt at a Solution

L = (N* magnetic flux)/i = (N*B*A*cos theta)/i = (n*l*A*B*cos theta)/(i), where N = n*l (l = wire length)

L2 = [(l/2)*A*(n/2)*B]/i = (1/4)*[(L*A*n*B)/i] = (1/4)*L??

Homework Statement

How much does the energy stored in an inductor change if the current through the inductor is doubled?

a. It remains the same.
b. It is doubled.
c. It is quadrupled.
d. It is halved.
e. It is quartered.

Homework Equations

See below.

The Attempt at a Solution

U = 0.5*L*I^2

U2 = 0.5*L*(2I)^2 = 4*0.5*L*I^2 = 4*U??

Homework Statement

The magnetic flux through a loop is made to vary according to the relation magnetic flux = 6t^2 + 7t + 1, where the units are SI. The emf induced in the loop when t = 2 s is

a. 38 V
b. 39 V
c. 40 V
d. 31 V
e. 19 V

Homework Equations

Emf_induced = -(d magnetic flux)/(dt)

The Attempt at a Solution

Taking the derivative, |emf| = 12t + 7, where t = 2 s

|Emf| = 12*(2s) + 7 = 31 V??

Thanks.

 

[Mentz114]

You don't seem to be having any problems with this. They all look correct.

 

[m2003]

Your solutions for the first two problems are correct. For the third problem, you have correctly calculated the magnitude of the induced emf, but you forgot to include the negative sign. The correct answer would be -31 V. This is because the direction of the induced emf is opposite to the change in magnetic flux, which in this case is decreasing.