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kvv77 [185]
3 years ago
5

2. The magnetic field inside a 4.0 cm diameter superconducting solenoid varies sinusoidally between 8.0 T and 12.0 T at a freque

ncy of 10 Hz. a. What is the maximum electric field strength at a point 1.5 cm from the solenoid axis?
Physics
1 answer:
Step2247 [10]3 years ago
3 0

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}}

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