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leva [86]
3 years ago
12

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]3 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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The answer is "\bold{9600 \frac{N}{m^3}, 978.59 \frac{kg}{m^3}, and \ 0.978}"

Explanation:

Given:

\to v=1\ liter= 10^{-3} \ m^3\\\\\to  w= 9.6 \ N\\

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Answer:

The average induced emf in the coil is 0.0286 V

Explanation:

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time of change in magnetic field, t = 0.1 s

The induced emf in the coil is calculated as;

E = A(dB)/dt

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A is area of the coil = πr²

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E = -0.00985(B₂-B₁)/t

E = 0.00985(B₁-B₂)/t

E = 0.00985(0.53 - 0.24)/0.1

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Therefore, the average induced emf in the coil is 0.0286 V

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