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:
2.65m/s
Explanation:
Using the equation of motion:
v² = u²+2a∆S where
v is the final velocity
u is the initial velocity
∆S is the change in distance
a is the acceleration
Given
u = 0m/s
a = 9.8m/s²
∆S = 1.3-0.943
∆S = 0.357m
Substituting the given parameters into the formula
v² = 0²+2(9.8)(0.357)
v² = 0+6.9972
v² = 6.9972
v=√6.9972
v = 2.65m/s
Hence the velocity at which it hit the ground is 2.65m/s
Answer: a) 6.67cm/s b) 1/2
Explanation:
According to law of conservation of momentum, the momentum of the bodies before collision is equal to the momentum of the bodies after collision. Since the second body was initially at rest this means the initial velocity of the body is "zero".
Let m1 and m2 be the masses of the bodies
u1 and u2 be their velocities respectively
m1 = 5.0g m2 = 10.0g u1 = 20.0cm/s u2 = 0cm/s
Since momentum = mass × velocity
The conservation of momentum of the body will be
m1u1 + m2u2 = (m1+m2)v
Note that the body will move with a common velocity (v) after collision which will serve as the velocity of each object after collision.
5(20) + 10(0) = (5+10)v
100 + 0 = 15v
v = 100/15
v = 6.67cm/s
Therefore the velocity of each object after the collision is 6.67cm/s
b) kinectic energy of the 10.0g object will be 1/2MV²
= 1/2×10×6.67²
= 222.44Joules
kinectic energy of the 5.0g object will be 1/2MV²
= 1/2×5×6.67²
= 222.44Joules
= 111.22Joules
Fraction of the initial kinetic transferred to the 10g object will be
111.22/222.44
= 1/2
Answer:
Explanation:
From the question we are told that:
Number of turns 
Conductor each with side length 
Current 
Magnetic field 
Generally the equation for the total magnetic moment M is mathematically given by
