A bar magnet of length 10 cm and pole strength 20 A-m is deflected through 30 degrees from the magnetic meridian. If earth horizontal component field is 320/4pie A-m .the moment of the deflecting couple is?

Question

Answer

Answer

Moment of the Deflecting Couple in a Bar Magnet

Given:
Length of the bar magnet (l) = 10 cm = 0.1 m
Pole strength (m) = 20 A-m
Deflection angle (θ) = 30 degrees
Horizontal component of the Earth's magnetic field (Bh) = 320/(4π) A-m

To find:
Moment of the deflecting couple (M)

Formula:
Moment of the deflecting couple (M) = mBl sin(θ)

Explanation:

1. Conversion of length:
Given length of the bar magnet is 10 cm. We need to convert it to meters by dividing it by 100.
l = 10 cm = 0.1 m

2. Calculation of moment of the deflecting couple:
Using the formula mentioned above,
M = mBl sin(θ)

3. Calculation of magnetic field (B):
The magnetic field due to the Earth's magnetic field acts perpendicular to the magnetic meridian. It is given by Bh = 320/(4π) A-m.

4. Calculation of the moment of the deflecting couple:
M = mBl sin(θ)
= (20 A-m) * (0.1 m) * (320/(4π) A-m) * sin(30 degrees)
= 20 * 0.1 * (320/(4π)) * 0.5
= 0.5 * 20 * 0.1 * (320/(4π))
= 1 * 0.1 * (320/(4π))
= 32/(4π)
= 8/π
≈ 8/3.14
≈ 2.55 A-m²

Therefore, the moment of the deflecting couple is approximately 2.55 A-m².

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