Question

2. The magnetic field inside a 4.0 cm diameter superconducting solenoid varies sinusoidally between 8.0 T and 12.0 T at a frequency of 10 Hz. a. What is the maximum electric field strength at a point 1.5 cm from the solenoid axis?

Answer 1

Answer:

The maximum electric field strength is  $94.2 \times {10^{ - 2}}\;{\rm{V/m}}$

Explanation:

Substitute 4.0 cm for D

r = D / 2

r = 4 / 2 = 2cm

= 2 × 10⁻²m

Now, according to the question, the expression for the magnetic field is as follows:

B = 10T + 2T sin(2π(10Hz) t )

Now, calculate $\frac{{dB}}{{dt}}$

$\begin{array}{c}\\\frac{{dB}}{{dt}} = \frac{{d\left( {10{\rm{ T + 2 Tsin}}\left( {2\pi \left( {10{\rm{ Hz}}} \right)t} \right)} \right)}}{{dt}}\\\\ = \left( {40\pi {\rm{ T/s}}} \right)\cos 2\pi \left( {10{\rm{ Hz}}} \right)t\\\end{array}$

Therefore, the electric field expression is as follows:

$E = \left( {\frac{r}{2}} \right)\left( {40\pi {\rm{ T/s}}} \right)\cos 2\pi \left( {10{\rm{ Hz}}} \right)t$

Substitute 0 for t in the above equation

$E = \left( {\frac{r}{2}} \right)\left( {40\pi {\rm{ T/s}}} \right)\cos 2\pi \left( {10{\rm{ Hz}}} \right)(0)$

$E = \left( {\frac{r}{2}} \right)\left( {40\pi)$

Now substitute 1.5 cm for r in the above equation

$E = (\frac{0.015}{2} )40\times3.14\\\\= 94.2\times 10^-^2V/m$

The maximum electric field strength is  $94.2 \times {10^{ - 2}}\;{\rm{V/m}}$