Question

A long solenoid with 1.65 103 turns per meter and radius 2.00 cm carries an oscillating current I = 6.00 sin 90πt, where I is in amperes and t is in seconds. What is the electric field induced at a radius r = 1.00 cm from the axis of the solenoid? (Use the following as necessary: t. Let E be measured in millivolts/meter and t be measured in seconds.)

Answer 1

Answer:

The  electric field  is $35\cos(90\pi t)\ mV/m$

Explanation:

Given that,

Radius = 2.00 cm

Number of turns per unit length $n= 1.65\times10^{3}$

Current $I = 6.00\sin 90\pi t$

We need to calculate the induced emf

$\epsilon =\mu_{0}nA\dfrac{dI}{dt}$

Where, n = number of turns per unit length

A = area of cross section

$\dfrac{dI}{dt}$=rate of current

Formula of electric field is defined as,

$E=\dfrac{\epsilon}{2\pi r}$

Where, r = radius

Put the value of emf in equation (I)

$E=\dfrac{\mu_{0}nA\dfrac{dI}{dt}}{2\pi r}$....(II)

We need to calculate the rate of current

$I=6.00\sin 90\pi t$....(III)

On differentiating equation (III)

$\dfrac{dI}{dt}=90\pi\times6.00\cos(90\pi t)$

Now, put the value of rate of current in equation (II)

$E=\dfrac{4\pi\times10^{-7}\times1.65\times10^{3}\times\pi\times(2.00\times10^{-2})^2\times90\pi\times6.00\cos(90\pi t)}{2\pi\times 2.00\times10^{-2}}$

$E=35\cos(90\pi t)\ mV/m$

Hence, The  electric field  is $35\cos(90\pi t)\ mV/m$

Answer 2

Answer:

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Explanation: