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3 years ago

14

12. A flat circular coil of wire having 200 turns and diameter 6.0 cm carries a current of 7.0 A. It is placed in a magnetic fie

ld of with the plane of the coil making an angle of 30° with the magnetic field. What is the magnitude of the magnetic torque on the coil?

1 answer:

IgorLugansk [536]3 years ago

4 0

Answer:

The magnitude of the magnetic torque on the coil is 1.98 A.m²

Explanation:

Magnitude of magnetic torque in a flat circular coil is given as;

τ = NIASinθ

where;

N is the number of turns of the coil

I is the current in the coil

A is the area of the coil

θ is the angle of inclination of the coil and magnetic field

Given'

Number of turns, N = 200

Current, I = 7.0 A

Angle of inclination, θ = 30°

Diameter, d = 6 cm = 0.06 m

A = πd²/4 = π(0.06)²/4 = 0.002828 m²

τ = NIASinθ

τ = 200 x 7 x 0.002828 x Sin30

τ = 1.98 A.m²

Therefore, the magnitude of the magnetic torque on the coil is 1.98 A.m²

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Answer:

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Explanation:

The electrical reactance is defined as:

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$X_c=\frac{1}{2\pi *100*(47*10^{-6} )} =-33.86275385\Omega$

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Replacing the data:

$Z=100-j33.86275385$

In order to find the magnitude of the impedance we can use the next equation:

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We can use Ohm's law to find the current:

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And its magnitude is:

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Answer:

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or

1m = 1*10^(-3) km

if we replace the meter in the first equation, we get:

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1 μm = 1*10^(-6 - 3)km

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And for width, we have 47.66um, this is 46.66 times 1*10^(-9)km, or:

W = 46.66*1*10^(-9)km = 4.666*10^(-8) km

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