The magnitude of the magnetic dipole moment of the bar magnet is 1.2 Am²
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Magnetic dipole moment of the bar magnet</h3>
The magnitude of the magnetic dipole moment of the bar magnet at distance from its axis is calculated as follows;

where;
- B is magnetic field
- m is dipole moment
- μ is permeability of free space
m = (4π x 0.1³ x 2.4 x 10⁻⁴)/(2 x 4π x 10⁻⁷)
m = 1.2 Am²
The complete question is below:
What is the magnitude of the magnetic dipole moment of the bar magnet from 0.1 m of its axis and magnetic field strength of 2.4 x 10⁻⁴ T.
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Given: Heat Qout means useful work = 2800 J
Heat Qin = 8900 J
Required; Efficiency = ?
Formula: Efficiency = Qout/Qin = x 100%
= 2800 J/8900 J = 0.3146 X 100 %
Efficiency = 31.46%
Answer:
Derivation of Conservation of Momentum
Applying Newton's third law, these two impulsive forces are equal and opposite i.e. is equal to the change in momentum of the first object. is equal to the change in momentum of the second object. This relation suggests that momentum is conserved during the collision.
Explanation:
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Chemical change is two things get chemical reaction and produce a new product.