Answer:
The maximum electric field strength is [tex]94.2 \times {10^{ - 2}}\;{\rm{V/m}}[/tex]
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 [tex]\frac{{dB}}{{dt}}[/tex]
[tex]\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}[/tex]
Therefore, the electric field expression is as follows:
[tex]E = \left( {\frac{r}{2}} \right)\left( {40\pi {\rm{ T/s}}} \right)\cos 2\pi \left( {10{\rm{ Hz}}} \right)t[/tex]
Substitute 0 for t in the above equation
[tex]E = \left( {\frac{r}{2}} \right)\left( {40\pi {\rm{ T/s}}} \right)\cos 2\pi \left( {10{\rm{ Hz}}} \right)(0)[/tex]
[tex]E = \left( {\frac{r}{2}} \right)\left( {40\pi)[/tex]
Now substitute 1.5 cm for r in the above equation
[tex]E = (\frac{0.015}{2} )40\times3.14\\\\= 94.2\times 10^-^2V/m[/tex]
The maximum electric field strength is [tex]94.2 \times {10^{ - 2}}\;{\rm{V/m}}[/tex]
The maximum electric field strength at a point 1.5 cm from the solenoid axis is; E_max = 0.942 V/m
We are given;
Diameter; d = 4 cm = 0.04 m
Radius; r = d/2 = 0.04/2 = 0.02 m
magnetic field varies sinusoidally between 8T and 12 T.
Frequency; f = 10 Hz
From Faraday’s law, the electric field is given by;
E = -(r/2)(dB/dt)
From the variation of the magnetic field, we can write the equation of the magnetic field as;
B(t) = B_c + B_o*sin (2πft)
B(t) = 10 + 2sin (2π * 10t)
Thus;
dB/dt = 40π cos 20πt
Maximum value of electric field will occur when cos 20πt = -1. Thus;
(dB/dt)_max = -40π
Thus at r = 1.5 cm = 0.015 m from the solenoid axis, we have;
E = -(0.015/2) × -40π
E = 0.942 V/m
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