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.
Learn more about dipole moment here: brainly.com/question/27590192
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Answer:
6.8 m/s2
Explanation:
Let g = 9.8 m/s2. The total weight of both the rope and the mouse-robot is
W = Mg + mg = 1*9.8 + 2*9.8 = 29.4 N
For the rope to fails, the robot must act a force on the rope with an additional magnitude of 43 - 29.4 = 13.6 N. This force is generated by the robot itself when it's pulling itself up at an acceleration of
a = F/m = 13.6 / 2 = 6.8 m/s2
So the minimum magnitude of the acceleration would be 6.8 m/s2 for the rope to fail
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