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2 years ago

A solenoid of length 10 cm has 247 turns and a magnetic field at the center of 1.2 T. What is the current through the solenoid?

Physics

1 answer:

elena55 [62]2 years ago

6 0

Answer:

386.6 A

Explanation:

We are given;

Solenoid length; L = 10 cm = 0.1 m

Number of turns; N = 247

Magnetic field; B = 1.2 T

Now, at the centre of a long solenoid of N turns/metre carrying a current(I), the formula for the magnetic field is given as;

B = μ_oNI/L

Making I the subject we have;

I = BL/μ_oN

Where μ_o is the vacuum of permeability and has a constant value of 4π × 10^(−7) H/m

Thus;

I = (1.2 × 0.1)/(4π × 10^(−7) × 247)

I = 386.6 A

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2 years ago

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

Leno4ka [110]

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)

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