Question

The induced electric field 13 cm from the axis of a solenoid with a 9.0 cm radius is 44 V/m.

Find the rate of change of the solenoid's magnetic field.

Answer 1

Concepts and reason

The concepts required solve the given problem are Faraday’s law of induction, and definition of magnetic flux.

Initially, using the Faraday’s law and magnetic flux expression find out the equation for the rate of change of magnetic field.

Then find the induced ${\rm{emf}}$ in the solenoid by using expression for induced ${\rm{emf}}$ in terms of electric field.

Fundamentals

The magnetic flux through a closed loop of current carrying wire can be given by the following expression:

$\phi = \vec B.\vec A$

Here, B is the magnetic field and A is the area of the loop.

The expression of the emf generated within a solenoid due to the change in magnetic flux passing through it is as follows:

$\varepsilon = - \frac{{d\phi }}{{dt}}$

Here, $\frac{{d\phi }}{{dt}}$ is the rate of change of magnetic flux through the solenoid with respect to time.

The expression of the induced ${\rm{emf}}$ in terms of electric field is give as follows:

$\varepsilon = - \int {E \cdot dl}$

According to Faraday’s law, the induced ${\rm{emf}}$ is given as follows:

$\varepsilon = - \frac{{d\phi }}{{dt}}$

Substitute $BA$ for $\phi$ in the above equation.

$\begin{array}{c}\\\varepsilon = - \frac{{d\left( {BA} \right)}}{{dt}}\\\\ = - A\frac{{dB}}{{dt}}\\\end{array}$

Rearrange the above equation for change in magnetic field.

$\frac{{dB}}{{dt}} = - \frac{\varepsilon }{A}$

The ${\rm{emf}}$ induced in terms of electric field is give as follows:

$\begin{array}{c}\\\varepsilon = - \int {E \cdot dl} \\\\ = - E\int {dl} \\\end{array}$

The integral $\int {dl}$ value is equal to $2\pi r.$

Substitute $2\pi r$ for $\int {dl}$ in the above equation.

$\begin{array}{c}\\\varepsilon = - E\left( {2\pi r} \right)\\\\ = - 2E\pi r\\\end{array}$

Substitute $- 2E\pi r$ for $\varepsilon$ in the above equation $\frac{{dB}}{{dt}} = - \frac{\varepsilon }{A}.$

$\begin{array}{c}\\\frac{{dB}}{{dt}} = - \frac{{\left( { - 2E\pi r} \right)}}{A}\\\\ = \frac{{2E\pi r}}{A}\\\end{array}$

The cross section area A is give as followed:

$A = \pi {R^2}$

Substitute $\pi {R^2}$ for A in the above equation.

$\begin{array}{c}\\\frac{{dB}}{{dt}} = \frac{{2E\pi r}}{{\pi {R^2}}}\\\\ = \frac{{2Er}}{{{R^2}}}\\\end{array}$

Substitute 44 V/m for E, 9.0 cm for R, and 13 cm for r in the above equation.

$\begin{array}{c}\\\frac{{dB}}{{dt}} = \frac{{2\left( {44\,{\rm{V/m}}} \right)\left( {13\;{\rm{cm}}} \right)\left( {\frac{{1\,{\rm{m}}}}{{100\,{\rm{cm}}}}} \right)}}{{{{\left( {9.0\,{\rm{cm}}} \right)}^2}\left( {\frac{{1\;{{\rm{m}}^2}}}{{{{10}^4}\,{\rm{c}}{{\rm{m}}^2}}}} \right)}}\\\\ = 1412\,{\rm{T/s}}\\\end{array}$

Ans:

The rate of change of solenoid’s magnetic field is 1412 T/s.