- #1

gsmtiger18

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## Homework Statement

A long straight wire has a line charge, λ that varies in time according to: λ = λ

_{0}e

^{(-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 = λ

_{0}e

^{(-βt)}

I = dq/dt

B = μ

_{0}I/2πr

displacement current: i = ∈

_{0}dΦ

_{e}/dt

Magnetic flux: Φ

_{m}= ∫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 = μ

_{0}dI/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.

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