For a solar eclipse to occur which of the following alignments must be necessary
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The magnitude of the magnetic dipole moment of the bar magnet is 1.2 Am²
<h3> 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:
a)N = 3.125 * 10¹¹
b) I(avg) = 2.5 × 10⁻⁵A
c)P(avg) = 1250W
d)P = 2.5 × 10⁷W
Explanation:
Given that,
pulse current is 0.50 A
duration of pulse Δt = 0.1 × 10⁻⁶s
a) The number of particles equal to the amount of charge in a single pulse divided by the charge of a single particles
N = Δq/e
charge is given by Δq = IΔt
so,
N = IΔt / e
N = 3.125 * 10¹¹
b) Q = nqt
where q is the charge of 1puse
n = number of pulse
the average current is given as I(avg) = Q/t
I(avg) = nq
I(avg) = nIΔt
= (500)(0.5)(0.1 × 10⁻⁶)
= 2.5 × 10⁻⁵A
C) If the electrons are accelerated to an energy of 50 MeV, the acceleration voltage must,
eV = K
V = K/e
the power is given by
P = IV
P(avg) = I(avg)K / e
= 1250W
d) Final peak=
P= Ik/e
=
P = 2.5 × 10⁷W