> #electrical-engineering , #physics > Given N=20, diameter=8cm, current=3.0A, magnetic moment=0.0015 If coil is suspended in uniform magnetic field Weber/m^2 calculate the moment of couple acting on it?
Anonymous
Sep 4

Given N=20, diameter=8cm, current=3.0A, magnetic moment=0.0015 If coil is suspended in uniform magnetic field Weber/m^2 calculate the moment of couple acting on it?

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C
ChatGPT
Sep 4
To calculate the moment of couple acting on the coil, we can use the formula:

Moment of couple = magnetic moment * magnetic field strength * sin(theta)

Given:
- Magnetic moment = 0.0015
- Diameter of the coil = 8 cm = 0.08 m (radius = 0.04 m)
- Current = 3.0 A
- Number of turns = N = 20
- Area of the coil = pi * r^2 = pi * (0.04)^2 = 0.005 m^2

The magnetic field strength can be calculated using the formula:

B = u0 * N * I / (2 * r)

Where:
- u0 = permeability of free space = 4 * pi * 10^-7 Tm/A
- N = number of turns = 20
- I = current = 3.0 A
- r = radius of the coil = 0.04 m

B = (4 * pi * 10^-7) * 20 * 3.0 / (2 * 0.04)
B = 0.075 T

Now, we can calculate the moment of couple:

Moment of couple = 0.0015 * 0.075 * sin(theta)

Since the coil is suspended in a uniform magnetic field, the angle between the magnetic moment and the magnetic field is 90 degrees. Therefore, sin(theta) = 1.

Moment of couple = 0.0015 * 0.075 * 1
Moment of couple = 0.0001125 Nm

Therefore, the moment of couple acting on the coil is 0.0001125 Nm.
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G
Google Gemini AI
Sep 4
Given:

- Number of turns: N = 20
- Diameter: d = 8 cm = 0.08 m
- Current: I = 3.0 A
- Magnetic moment: M = 0.0015 Am^2
- Magnetic field strength: B = ? (uniform)

To find:

- Moment of couple: Γ = ?

Formula:

The moment of couple acting on a current loop in a magnetic field is given by:

Γ = NIAB sin θ

where:

- N is the number of turns in the coil
- I is the current in the coil
- A is the area of the coil
- B is the magnetic field strength
- θ is the angle between the magnetic field and the normal to the coil

Calculation:

First, we need to find the area of the coil:

A = πd^2/4 = π(0.08 m)^2/4 = 0.005027 m^2

Now, substituting the given values into the formula for the moment of couple:

Γ = NIAB sin θ = (20)(3.0 A)(0.005027 m^2)(B) sin 90°

Since θ = 90°, sin 90° = 1. Therefore:

Γ = (20)(3.0 A)(0.005027 m^2)(B)

Γ = 0.3016B Nm

Conclusion:

The moment of couple acting on the coil is 0.3016B Nm. Since the magnetic field strength B is not given, we cannot determine the exact value of the moment of couple.
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