A very long, straight solenoid with a cross-sectional area of 2.00 cm^2 is wound with 90.0 turn

A very long, straight solenoid with a cross-sectional area...

A very long, straight solenoid with a cross-sectional area of $2.00 \mathrm{~cm}^{2}$ is wound with 90.0 turns of wire per centimeter. Starting at $t=0$ the current in the solenoid is increasing according to $i(t)=\left(0.160 \mathrm{~A} / \mathrm{s}^{2}\right) t^{2}$. 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 $3.20 \mathrm{~A}$ ?

Key Concepts

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 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.

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 The given problem, cross -sectional area of the very long straight solenoid is a is equal to 2 .00 centimeter square or we can say this is 2 .00 into 10 dashed the power minus 4 meter square.

00:32 Number of turns per unit length in it is and is equal to 90 .0 turns per centimeter or we can say 90 in 200 means 9 ,000 turns per meter.

00:59 The current passing through this solenoid is varying as a function of time, which is given as i t is equal to 0 .160 t squared making a time derivative of this function is di by ttt means the time date of change of current will be given as 0 .160 twice of the time means it becomes 0 .32 now it is given that then the current is i is equal to 3 .20 ampere then we have to find this time rate of change no we have to find the time when the current is 3 .20 ampere so to find it we put the value of current here in equation number one so in place of this current we use 3 .20 is equal to 0 .160 t squared.

02:31 So the function of time comes out to be t squared is equal to 3 .20 divided by 0 .160 which becomes 20.

02:46 So time is given as square root of 20 seconds, which comes out to be, or we can leave it as it is only, square root of 20 seconds.

03:02 So now the time rate of change of current passing through the solenoid is given as, as we know, this is given as 0 .3 times of this time.

03:16 So in place of t, you can use square root of 20, ampere per second.

03:24 So following this, it comes out to be 1 .43 amper second.

03:31 And this is the time rate of change of current passing through the solenoid.

03:38 Now, the magnetic field at the axis of this solenoid will be given by mu not into n1i.

03:54 Or we can say here this n1 is the number of turns per unit length, which is given to us as 9 ,000 turns per meter.

04:07 Using this expression, we can find the magnetic flux.

04:13 This is the magnetic field linked through the primary solenoid.

04:17 So the magnetic flux linked through the secondary coil, which is having total five terms in it...

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