Magnetism and Matter Test 3

According to recent researches the magnetic field of earth is considered due to a large bar magnet situated in earth's core.
It is considered that the north pole of this large magnet is situated at the geographical south of earth and vice versa and as the magnetic field due to a bar magnet is from north pole to south pole of the maget thus the earth's magnetic field is considered from geographical south to geographical north which are respectively north and south poles of the bar magnet.

μ = μ₀μ_r. μ is the permeability of medium, μ₀ is permeability of free space, and μr is relative permeability of the medium.

B=B₀+Bm
= μ₀nI+ μmnI
=nI(μ+ μm)
=60x5(5000μ₀+ μ₀)
B=300(5001μ₀)
B=1.88T
H=B/μm
=300x5001μ₀/5000μ
=300AT

Work required to turn the dipole is given by
W=MB[cosθ_i- cosθ_f]
Here θi is the initial angle made by a dipole with a magnetic field and   is the final angle made by a dipole with a magnetic field.
magnetic moment is normal to the field direction.
so, θ_i=0 and θ_f=90
W = 1.5 × 0.22 [ cos0° - cos90° ]
W = 0.33J
 
magnetic moment is opposite to the field direction.
so, θ_i=0 and θ_f=180
now, W = 1.5 × 0.22 [ cos0° - cos180°]
= 0.33 [ 1 - (-1) ]
= 0.66 J

The earth's magnetic field is Be​ and its horizontal and vertical components are H_e​ and H_v​
cosθ= H_e​​/B_e​
∴cos60^o= (​0.26×10^−4​/ B_e )T
⇒B_e​=(​0.26×10⁻⁴)/ (½)​=0.52×10⁻⁴T=0.52G

Explanation:As magnetic monopole does not exist. If we split the bar magnet into two pieces  each part will have its own north and south pole.

All substances show some kind of magnetic behaviour. After all, they are made up of charged particles: electrons and protons. It is the way in which electron clouds arrange themselves in atoms and how groups of these atoms behave that determines the magnetic properties of the material. The atom (or group of atoms) in effect becomes a magnetic dipole or a mini bar magnet that can align according to the magnetic field applied. The net effect of all these dipoles determines the magnetic properties of the Magnetic Materials.

Correct Answer :- C

Explanation : Magnetic Moment = Magnetization × Volume

Side of cube = 20 cm

= 0.2 m

= 500 × 0.2 × 0.2 × 0.2

= 4Am²

 =1.5x0.22
 =15x22x10-3
 =330x10-3
Τ=0.33J
 

At null point (and along the axis), earth's magnetic field and bar's magnetic field are opposite in direction.
On the equatorial line, the bar's magnetic field is opposite in direction to its field on the axis. Hence, on the equatorial line, the two fields add up.
As the null points are on the axis of the bar magnet, therefore,
B1= (μ0/4π)(2M/d3)=H
On the equatorial line of magnet at same distance(d), field due to the magnet is 
B2= (μ0/4π)(M/d3)=B1/2=H/2
Therefore, total magnetic field at this point on equatorial line is 
B=B2+H=3H/2=3×0.36G/2=0.54G
 

Explanation:North pole act as positive end and South pole act as negative end. The iron fillings in the presence of magnetic field  gets magnetized and form a tiny magnet which then gets  attracted to the poles of bar magnet. North pole of bar magnet attracts south pole of tiny magnet and vice-versa. So all the iron fillings are arranged as the magnetic field lines of bar magnet. 

When a diamagnetic material is placed in an external magnetic field the spin motion of electrons is so modified that the electrons which produce the moments in the direction of the external field show down while the electrons which produce magnetic moments in opposite directions get accelerated.
Thus, a net magnetic moment is induced in the opposite direction of the applied magnetic field. Hence the substance is magnetized opposite of the external field. Thus, it moves from stronger. Weaker parts of the magnetic.
 

Correct Answer :- d

Explanation : μ_total = MV

= (8*10⁵)(2*10^-2)

= 6A m²

Magnetic field on the axis of a current loop with magnetic material μtotal is:

B = μ₀ μ_total / (2π(x² + a²)^1/2

B = (4π*10^-7)(6) / [2π(0.1)]

= 1 * 10^-3 T

= 0.001 T

 Answer :- 

Solution :- torque when =180°

= 1.5 × 0.22 × sin 180°

= 0.33 × 0 = 0 Nm

According to the question, the resultant field is inclined at 45o with earth's magnetic field.
tanθ=B_H​​/B
θ=45^o
tanθ=1=B_H​​/B
B_H​=B= μ_o​M​/4πr³
Given, B= 0.42×10⁻⁴T
M=5.25×10⁻²J/T
Therefore,0.42+10⁻⁴=10⁻⁷×[(5.25×10⁻²)/r³]​
r³=12.5×10⁻⁵=125×10⁻⁶
r=5×10⁻²m = 5 cm

Magnetic poles exist in pairs. It is not possible to isolate a north pole or a south pole. Magnetic field lines start from the north pole and go to the south pole and return to the north pole. They form continuous closed loops unlike electric lines of force which do not as an electric monopole, a single charge does exist.

It is because when we make a group of electrons inside the material it is called domain. So when we apply an external magnetic field on the material. Those domains start to align themselves in the direction of the magnetic field. BUT NOT ALL DOMAINS ALIGN THEMSELVES IN THAT DIRECTION. So option D is correct.

When a parametric material comes close to a magnetic field the atoms will align with the magnetic field which causes another magnetic field. Permeability is the total magnetic field divided by the original magnetic field. Increase in temperature makes it harder for the atoms to align which decreases the strength of the second field. This results in a lower total magnetic field (numerator) and thus lower permeability.

We know the horizontal component of earth magnetic field H=R cos θ
Where θ=angle of dip=22⁰ and the value of H=0.35G
So, putting the values respectively R= H/cosθ=0.35/cos22=0.35/0.927=0.377=0.38G

Given data,
Rowland ring is toroid with a core of magnetic material,
Radius of the ring r=15cm
Number of turns of the wire N=3500
Relative permeability of the core,μ_r​=800
Magnetising current, I=1.2A
Magnetic field, B=?
Using the formula for the magnetic field of a toroid,
B=μ_0​nI
Number of turns per unit length, n=N/2πr=3500/​(2π×15×10⁻²)
Therefore,
B=μ_0​μ_r​nI
The magnetic field is given by the relation,
B= μr​μ0​IN​/2πr
= (800×4π×10⁻⁷×1.2×3500)/​ 2π×0.15 =4.48T

A bar magnet may be thought of as a large number of circulating currents in analogy with a solenoid. On cutting a bar magnet in half we get two smaller solenoids with weaker magnetic properties.The field lines remain continuous, emerging from one face of the solenoid and entering into the other face. (i.e. similar magnetic fields). The magnetic moment of a bar magnet is also equal to the magnetic moment of an equivalent solenoid.
 

For paramagnetic materials orbital and spin magnetic moments of the electrons are of the order of bohr magneton.
Paramagnetic materials have some unpaired electrons due to these unpaired electrons the net magnetic moment of all electrons in an atom is not added up to zero. Hence atomic dipole exists in this case. On applying external magnetic field the atomic dipole aligns in the direction of the applied external magnetic field. In this way, paramagnetic materials are feebly magnetized in the direction of the magnetizing field.

If a magnet is suspended over a container of liquid air, it attracts droplets to its poles. The droplets contain only liquid oxygen; even though nitrogen is the primary constituent of air, it is not attracted to the magnet. Explain what this tells you about the magnetic susceptibilities of oxygen and nitrogen, and explain why a magnet in ordinary, room-temperature air doesn’t attract molecules of oxygen gas to its poles.

Here M=0.48JT⁻¹,d=10cm=0.1m
On axis,
B= (μ_0​​/4π)(2M/d3)​=10⁻⁷×(2×0.48/0.1³)​=0.96×10⁻⁴T

The needle will point along W-E if the result of the earth’s magnetic field and magnetic field due to the coil have a resultant in the W-E direction. This happens if B cos 45^o=Earth's field
⟹ (μ0​nI​/2r) cos45^o=(4π×10−7×30×0.35​)/(2×0.12) ×(1/√2)​​=0.39G=Earth's magnetic field.

For a short Bar Magnet, the magnetic induction at a point on the axis at a distance r from centre is given by the formula
B= (μ0/4π​​) (2M​/r³)
⇒ B= K​/ r³
⇒ B∝1/r³​
The correct answer is option D.

A ferromagnetic substance contains permanent atomic magnetic dipoles that are spontaneously oriented parallel to one another even in the absence of an external field. The magnetic repulsion between two dipoles aligned side by side with their moments in the same direction makes it difficult to understand the phenomenon of ferromagnetism. It is known that within a ferromagnetic material, there is a spontaneous alignment of atoms in large clusters. A new type of interaction, a quantum mechanical effect known as the exchange interaction, is involved. A highly simplified description of how the exchange interaction aligns electrons in ferromagnetic materials is given here.

Spin and Orbital angular momentum arise due to the presence of unpaired electrons. If there are unpaired electrons, these momenta will be there otherwise not.

Given,
Number of turns, N = 500 turns
Radius of solenoid, r = 2.9
Relative permeability of annealed iron of Km=1400
Permeability of free space,  μ₀ =4π
The magnetic field, B=0.350 T
Therefore,
μ/μ₀= μ_r
μ= μ_r x μ₀
B= μ_rμ₀NI/2πR
0.350=1400x4x 3.14x500xI/2xπx2.90
I=72.mA
 

The Bohr magneton μ_B​ is a physical constant and the natural unit for expressing the magnetic moment of an electron caused by either its orbital or spin angular momentum.
μ_B​= eℏ​/2me ​
where e is the elementary charge, ℏ is the reduced Planck's constant, me​ is the electron rest mass.
The value of Bohr magneton in SI units is 9.27400968(20)×10⁻²⁴JT−1