Question

phys 252 laboratory 7 - magnetism introduction
lab introduction and purpose
in this lab you will conduct several experiments to explore the properties of permanent magnets and the magnetic fields they produce. you will also examine induced magnets and their properties. to prepare for this lab, you should review the following concepts from your lecture course:

• magnetism
• magnetic flux
• magnetic dipole moment μ
• gausss law for magnetism

in this lab, you will use a compass to map electric field lines. a compass is a small magnet with one pole marked in red to indicate
orth.\ this refers to geographical north... the north pole of the earth is actually a magnetic south pole! when you hold the compass away from a magnet, the north arrow will point toward the earths north pole. in lab this week you will work with small magnets that have magnetic fields much stronger than the earths at short distances. near these magnets, the compass arrow will align with the magnetic field produced by those small magnets. you will use this to draw the magnetic field lines in various different situations.
remember that the magnetic field of an ideal dipole measured along its axis (straight along the magnets central line from s to n) is given by: ( b_{axis}=\frac{mu_0}{4pi}\frac{2mu}{d^3} ) where permeability constant ( mu_0 = 4picdot10^{-7}n/a^2 )
the magnetic moment μ is a property of the magnet, and is expected to be constant.
pre - lab questions
please read through the full lab introduction and lab directions, then answer the following questions:

1. explain the meaning of magnetic flux in your own words. (2 pts)
2. what do you expect to see when you place a magnet near iron filings? (2 pts)
3. write the equation form of gausss law for magnetism, and explain what it means in your own words. how is it different from gausss law for electric flux? (2 pts)
4. describe how a compass works, and how you will use a compass in this lab. (2 pts)
5. solve the following equation for magnetic dipole moment μ: ( b_{axis}=\frac{mu_0}{4pi}\frac{2mu}{d^3} ) (2 pts)

Explanation:

Step1: Isolate the variable μ

Multiply both sides of the equation \(B_{axis}=\frac{\mu_{0}}{4\pi}\frac{2\mu}{d^{3}}\) by \(\frac{4\pi d^{3}}{\mu_{0}}\).

$$
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Step2: Simplify the equation

Simplify the right - hand side of the equation. \(\frac{\mu_{0}}{4\pi}\times\frac{4\pi d^{3}}{\mu_{0}}\times\frac{2\mu}{d^{3}} = 2\mu\). So, \(\mu=\frac{B_{axis}\times4\pi d^{3}}{2\mu_{0}}=\frac{2\pi B_{axis}d^{3}}{\mu_{0}}\)

Answer:

\(\mu=\frac{2\pi B_{axis}d^{3}}{\mu_{0}}\)