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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Answer:
87.1 mph
Explanation:
We are given that
Mass,m=60 kg
Power,P=340 W
Speed,v=5 m/s
Area,
Drag coefficient,
Coefficient of rolling resistance,
Friction force,
Where 
Let speed of cyclist=v'
Drag force,
Density of air,

Power,P=



1 m=0.00062137 miles
1 hour=3600 s
The best answer is B)
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Answer
given,
capacitance = C = 3.4-µF
inductance = L = 0.08 H
frequency is expressed as

time period

after time T/4 current reach maximum



t = 8.2 x 10⁻⁴ s
t = 0.82 ms
b) using law of conservation





I = 0.010 A
I = 10 mA
Answer:
E)brain decay
Explanation:
Looking at the question causes it.