# Displacement current and magnetic flux through a wire loop

• gsmtiger18
In summary, the conversation discusses the calculation of expressions for displacement current and magnetic flux through a square loop adjacent to a long straight wire with a varying line charge. The approach involves finding the current and magnetic field equations, and then using them to derive equations for the displacement current and magnetic flux. However, there may be some confusion about the orientation of the square loop in relation to the wire, and the potential for magnetic flux cancellation due to symmetry.
gsmtiger18

## Homework Statement

A long straight wire has a line charge, λ that varies in time according to: λ = λ0e(-βt). A square loop of dimension, a, is adjacent to the wire (at a distance a away from the wire). Calculate expressions for the displacement current at the center of the wire loop and the magnetic flux through the loop.

## Homework Equations

λ = λ0e(-βt)
I = dq/dt
B = μ0I/2πr
displacement current: i = ∈0e/dt
Magnetic flux: Φm = ∫A ⋅ dB

## The Attempt at a Solution

I know I can take the derivative of the charge to get current, which I need to find the magnetic field, which I need to find magnetic flux. However, I'm not sure what to do about the length component of λ. How can I get an expression for the charge without knowing a finite length? Once I get the current though, finding the flux and displacement current should be simple.

To get magnetic flux, I can first derive the current equation and then derive the equation for the magnetic field with respect to current to give dB = μ0dI/2πr. Then I can use the magnetic flux equation above to find an equation for the flux.

For the displacement current, I can use the relationship E = volts/meter, which I can manipulate to give E = IR/1.5a, where 1.5a is the distance to the center of the wire loop from the current-carrying wire. Taking the derivative with respect to current will give me an equation for dE, which I can use with the formula for the derivative of the electric flux. I can then use that with the formula for the displacement current to get an equation for the displacement current.For reference, the square wire loop is oriented with one side parallel to the line charge, at a distance a from the wire. This means that the sides of the square perpendicular to the wire go from distance a to 2a.

Last edited:
I'm no expert in this area, but you do not seem to be getting any replies, so...
I cannot understand why there would be a magnetic flux through the square with the wire and square coplanar. The electric field would be radial, and the displacement current density at the square would therefore be in the plane of the square (time derivative of ε0E). By symmetry, the magnetic flux would seem to cancel everywhere. Am I missing something?

## What is displacement current?

Displacement current is an electric current that is not the result of the flow of charge carriers, but rather the changing electric field in a region. It is an important concept in electromagnetism and is described by Maxwell's equations.

## What is the relationship between displacement current and magnetic flux?

According to Maxwell's equations, a changing electric field leads to a magnetic field, and a changing magnetic field leads to an electric field. Thus, displacement current is responsible for the generation of a magnetic field, which is measured as magnetic flux through a wire loop.

## How is magnetic flux through a wire loop calculated?

Magnetic flux through a wire loop is calculated by multiplying the strength of the magnetic field passing through the loop by the area of the loop, and then taking into account the angle between the magnetic field and the loop's surface.

## Why is displacement current important in electromagnetic induction?

Displacement current plays a crucial role in electromagnetic induction, which is the process of generating an electric current by changing the magnetic field passing through a wire loop. Without displacement current, electromagnetic induction would not be possible.

## Can displacement current be observed or measured directly?

No, displacement current cannot be observed or measured directly because it does not involve the flow of charge carriers. Instead, it is inferred from the changing electric and magnetic fields in a region, as described by Maxwell's equations.

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