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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?
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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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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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.