Rate of magnetic field change of a selnoid

Lance WIlliam · Sep 28, 2008

1. The problem statement, all variables and given/known data

The induced electric field 17cm from the axis of a solenoid with a 10 cm radius is 45V/m .
Find the rate of change of the solenoid's magnetic field?

dB/dt= T/s

2. Relevant equations
Faradays Law W/ the intergral due to changing position.

So [tex]\oint[/tex] E*dr =-d[tex]\phi[/tex](induced current)_B_/dt

3. The attempt at a solution

IM really not sure where to start...Advice on where to start would be VERY NICE!
Im also confused on V/m and a radii measure...not the meters but the "V".
Thankyou.

Gear300 · Sep 28, 2008

In this case the area and angle of the magnetic flux seems constant, so it should be the field that is changing. Whats happening is that the changing magnetic flux is propagating an electric field outward so that an electric field is induced across the solenoid. Since we're taking things in respect to the center axis of the solenoid, the electric field is constant throughout a circumference of equivalent radial distance. You can pull E out of the integral and integrate dr (the integration should yield the circumference).

Lance WIlliam · Sep 28, 2008

so once the E is pulled outside the intergral im left with "dr" to intergrate. "dr" being 45V/m? Im confused on the V/m as a measure for the radii. Why would I want the circumferance?
I guess what is comes down to is dB/dt....Is is saying Find the derivative of(magnetic flux / some measure of time)?

Gear300 · Sep 28, 2008

No, dr is increments of the path. Integrating dr gives the total length of the path, which is the circumference.

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Rate of magnetic field change of a selnoid

Lance WIlliam

Gear300

Lance WIlliam

Gear300

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