A very long, straight solenoid with a crosssectional area of...

A very long, straight solenoid with a crosssectional area of 2.00 cm$^2$ is wound with 90.0 turns of wire per centimeter. Starting at t = 0, the current in the solenoid is increasing according to $i(t) = (0.160 A/s^2)t^2$. A secondary winding of 5 turns encircles the solenoid at its center, such that the secondary winding has the same cross-sectional area as the solenoid. What is the magnitude of the emf induced in the secondary winding at the instant that the current in the solenoid is 3.20 A?

Key Concepts

Magnetic Field in a Solenoid

The magnetic field produced by a long, straight solenoid is given by the formula B = ?? n I, where ?? is the permeability of free space, n is the number of turns per unit length, and I is the current. This uniform field within the solenoid is essential for determining the magnetic flux through any area matching the solenoid’s cross section.

Mutual Inductance and Secondary Windings

Mutual inductance refers to the phenomenon where a change in current in one coil (the primary) induces an emf in another nearby coil (the secondary) that encloses the same magnetic flux. Here, the induced emf in the secondary winding is determined by multiplying the rate of change of flux by the number of turns in the secondary coil, as dictated by Faraday's law.

Time-Dependent Currents and Differentiation

When the current in the solenoid changes with time, it leads to a time-varying magnetic field and hence a varying magnetic flux. The induced emf is found by taking the time derivative of the magnetic flux, which often involves applying the chain rule to differentiate expressions that depend on time through functions like I(t).

Faraday's Law of Electromagnetic Induction

Faraday's law states that a changing magnetic flux through a circuit induces an electromotive force (emf) in the circuit. This principle is central to calculating the induced emf in coils or secondary windings when the magnetic environment changes with time, such as when a current in a nearby solenoid varies.

Magnetic Flux

Magnetic flux is defined as the product of the magnetic field and the area through which the field lines pass, considering the angle between the field and the normal to the area. In the context of a solenoid with a constant cross-sectional area, the flux is directly proportional to the magnetic field inside the solenoid.

Transcript

00:01 For this problem, on the topic of electromagnetic induction, we are told that a long straight solenoid with a given cross -sectional area and number of turns starts at t is equal to zero with the current increasing according to a function of time, which is given.

00:19 Now we have a secondary winding of five turns which encircles a solenoid at its center, and the second winding has the same cross -sectional area as the solenoid.

00:31 Find the magnitude of the emf induced in the second winding at the instant the current in the first is 3 .2 ampiers.

00:43 Now to solve this problem we know that the changing current in the solenoid will cause a changing magnetic field and hence changing flux through the second winding.

00:53 This changing flux will induce an emf in the secondary coil.

00:58 Now the magnetic field of the solenoid b we know is equal to mu not times n times i where mu not is a constant n is the number of turns per unit length and i is the current and the induced emf is epsilon and this is equal to the number of turns n multiplied by the magnitude of d phi d t the rate of change of magnetic flux so for we first need to find the magnetic field strength so b is equal to mu not times n times i.

01:41 And this is 4 pi times 10 to the minus 7.

01:48 And that's an s -i units of tesla meter per ampere times the number of turns per unit length, which is 90 times 10 to the power 2, or 900.

02:07 And that's per meter.

02:09 And the current is given by the expression 0 .16 ampiers per square second times t squared.

02:25 So if we simplify this, we get the magnetic field b to be 1 .81 times 10 to the minus 3 tesla's per square second times the time t squared.

02:40 So we have the magnetic field as a function of time.

02:44 Now the total flux through the second winding, so phi 1 -2 is equal to, and this is the magnetic flux, 5 times the magnetic field strength b times 2 times 10 to the minus 4 square meters.

03:16 So if we substitute b into this expression, we get the flux through the second winding to be 1 .81 times 10 to the minus 6...

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