Question 1

Q. 5.1. Answer the following questions regarding earth's magnetism:

(a) A vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth's magnetic field.

(b) The angle of dip at a location in southern India is about $18^{\circ}$. Would you expect a greater or smaller dip angle in Britain?

(c) If you made a map of magnetic fie

Accepted Answer

(a) The three independent quantities conventionally used for specifying earth's magnetic field are:

(i) Magnetic declination,

(ii) Angle of dip, and

(iii) Horizontal component of earth's magnetic field

(b) The angle of dip at a point depends on how far the point is located with respect to the North Pole or the South Pole. The angle of dip would be greater in Britain (it is about $70^{\circ}$ ) than in southern India because the location of Britain on the globe is closer to the magnetic North Pole.

(c) It is hypothetically considered that a huge bar magnet is dipped inside earth with its north pole near the geographic South Pole and its south pole near the geographic North Pole.

Magnetic field lines emanate from a magnetic north pole and terminate at a magnetic south pole. Hence, in a map depicting earth's magnetic field lines, the field lines at Melbourne, Australia would seem to come out of the ground.

(d) If a compass is located on the geomagnetic North Pole or South Pole, then the compass will be free to move in the horizontal plane while earth's field is exactly vertical to the magnetic poles. In such a case, the compass can point in any direction.

(e) Magnetic moment, $\mathrm{M}=8 	imes 10^{22} \mathrm{~J} \mathrm{~T}^{-1}$

Radius of earth, $\mathrm{r}=6.4 	imes 10^{6} \mathrm{~m}$

Magnetic field strength, $B=\frac{\mu_{0} M}{4 \pi r^{3}}$

Where,

$$

\begin{aligned}

& \mu_{0}=	ext { Permeability of free space }=4 \pi 	imes 10^{-7} \mathrm{Tm} \mathrm{~A}^{-1} \

& 	herefore B=\frac{4 \pi 	imes 10^{-7} 	imes 8 	imes 10^{22}}{4 \pi 	imes\left(6.4 	imes 10^{6}ight)^{3}}=0.3 \mathrm{G}

\end{aligned}

$$

This quantity is of the order of magnitude of the observed field on earth.

(f) Yes, there are several local poles on earth's surface oriented in different directions. A magnetised mineral deposit is an example of a local N-S pole.

Question 2

Q. 5.1. Answer the following questions:
(a) The earth's magnetic field varies from point to point in space.

Does it also change with time? If so, on what time scale does it change appreciably?
(b) The earth's core is known to contain iron. Yet geologists do not regard this as a source of the earth's magnetism. Why?
(c) The charged currents in the outer conducting regions of the earth's core are t

Accepted Answer

(a) Earth's magnetic field changes with time. It takes a few hundred years to change by an appreciable amount. The variation in earth's magnetic field with the time cannot be neglected.
(b)Earth's core contains molten iron. This form of iron is not ferromagnetic. Hence, this is not considered as a source of earth's magnetism.
(c)The radioactivity in earth's interior is the source of energy that sustains the currents in the outer conducting regions of earth's core. These charged currents are considered to be responsible for earth's magnetism.
(d)Earth reversed the direction of its field several times during its history of 4 to 5 billion years. These magnetic fields got weakly recorded in rocks during their solidification. One can get clues about the geomagnetic history from the analysis of this rock magnetism.
(e)Earth's field departs from its dipole shape substantially at large distances (greater than about $30,000 \mathrm{~km}$ ) because of the presence of the ionosphere. In this region, earth's field gets modified because of the field of single ions. While in motion, these ions produce the magnetic field associated with them.
(f)An extremely weak magnetic field can bend charged particles moving in a circle. This may not be noticeable for a large radius path. With reference to the gigantic interstellar space, the deflection can affect the passage of charged particles.

Question 3

Q. 5.3. A short bar magnet placed with its axis at $30^{\circ}$ with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to $4.5 	imes 10^{-2} \mathrm{~J}$. What is the magnitude of magnetic moment of the magnet?

Accepted Answer

Magnetic field strength, $B=0.25 \mathrm{~T}$

Torque on the bar magnet, $\mathrm{T}=4.5 	imes 10^{-2} \mathrm{~J}$

Angle between the bar magnet and the external magnetic field, $	heta=30^{\circ}$

Torque is related to magnetic moment ( M ) as:

$\mathrm{T}=\mathrm{MB} \sin 	heta$

$$

\begin{aligned}

herefore M & =\frac{T}{B \sin 	heta} \

& =\frac{4.5 	imes 10^{-2}}{0.25 	imes \sin 30^{\circ}}=0.36 \mathrm{~J} \mathrm{~T}^{-1}

\end{aligned}

$$

Hence, the magnetic moment of the magnet is $0.36 \mathrm{~J} \mathrm{~T}^{-1}$.

Question 4

Q. 5.4. A short bar magnet of magnetic moment $\mathrm{m}=0.32 \mathrm{~J} \mathrm{~T}^{-1}$ is placed in a uniform magnetic field of 0.15 T . If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?

Accepted Answer

Moment of the bar magnet, $\mathrm{M}=0.32 \mathrm{~J} \mathrm{~T}^{-1}$

External magnetic field, $\mathrm{B}=0.15 \mathrm{~T}$

(a)The bar magnet is aligned along the magnetic field. This system is considered as being in stable equilibrium. Hence, the angle $	heta$, between the bar magnet and the magnetic field is $0^{\circ}$.

Potential energy of the system $=-M B \cos 	heta$

$=-0.32 	imes 0.15 \cos 0^{\circ}$

$=-4.8 	imes 10^{-2} \mathrm{~J}$

(b)The bar magnet is oriented $180^{\circ}$ to the magnetic field. Hence, it is in unstable equilibrium. $	heta=180^{\circ}$

Potential energy $=-\mathrm{MB} \cos 	heta$

$=-0.32 	imes 0.15 \cos 180^{\circ}$

$=4.8 	imes 10^{-2} \mathrm{~J}$

Question 5

Q. 5.5. A closely wound solenoid of 800 turns and area of cross section $2.5 	imes 10^{-4} \mathrm{~m}^{2}$ carries a current of 3.0 A . Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?

Accepted Answer

Number of turns in the solenoid, $\mathrm{n}=800$

Area of cross-section, $\mathrm{A}=2.5 	imes 10^{-4} \mathrm{~m}^{2}$

Current in the solenoid, $\mathrm{I}=3.0 \mathrm{~A}$

A current-carrying solenoid behaves as a bar magnet because a magnetic field develops along its axis, i.e., along its length.

The magnetic moment associated with the given current-carrying solenoid is calculated as:

$\mathrm{M}=\mathrm{n}$ I A

$=800 	imes 3 	imes 2.5 	imes 10^{-4}$

$=0.6 \mathrm{~J} \mathrm{~T}^{-1}$

Question 6

Q. 5.6. If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of $30^{\circ}$ with the direction of applied field?

Accepted Answer

Magnetic field strength, $\mathrm{B}=0.25 \mathrm{~T}$

Magnetic moment, $\mathrm{M}=0.6 \mathrm{~T}^{-1}$

The angle $	heta$, between the axis of the solenoid and the direction of the applied field is $30^{\circ}$. Therefore, the torque acting on the solenoid is given as:

$$

\begin{aligned}

au & =M B \sin 	heta \

& =0.6 	imes 0.25 \sin 30^{\circ} \

& =7.5 	imes 10^{-2} \mathrm{~J}

\end{aligned}

$$

Question 7

Q. 5.7. A bar magnet of magnetic moment $1.5 \mathrm{~J} \mathrm{~T}^{-1}$ lies aligned with the direction of a uniform magnetic field of 0.22 T .

(a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction?

(b) What is the torque on the magnet in cases (i) and (ii)?

Accepted Answer

(a) Magnetic moment, $\mathrm{M}=1.5 \mathrm{~J} \mathrm{~T}^{-1}$

Magnetic field strength, $B=0.22 T$

(i) Initial angle between the axis and the magnetic field, $	heta_{1}=0^{\circ}$

Final angle between the axis and the magnetic field, $	heta_{2}=90^{\circ}$

The work required to make the magnetic moment normal to the direction of magnetic field is given as:

$$

\begin{aligned}

W & =-M B\left(\cos 	heta_{2}-\cos 	heta_{1}ight) \

& =-1.5 	imes 0.22\left(\cos 90^{\circ}-\cos 0^{\circ}ight) \

& =-0.33(0-1) \

& =0.33 \mathrm{~J}

\end{aligned}

$$

(ii) Initial angle between the axis and the magnetic field, $	heta_{1}=0^{\circ}$

Final angle between the axis and the magnetic field, $	heta_{2}=180^{\circ}$

The work required to make the magnetic moment opposite to the direction of magnetic field is given as:

$$

\begin{aligned}

W & =-M B\left(\cos 	heta_{2}-\cos 	heta_{1}ight) \

& =-1.5 	imes 0.22\left(\cos 180-\cos 0^{\circ}ight) \

& =-0.33(-1-1) \

& =0.66 \mathrm{~J}

\end{aligned}

$$

(b)For case_(i): $	heta=	heta_{2}=90^{\circ}$

$$

\begin{aligned}

& 	herefore 	ext { Torque, } \quad 	au=M B \sin 	heta \

& =1.5 	imes 0.22 \sin 90^{\circ} \

& =0.33 \mathrm{~J}

\end{aligned}

$$

For case (ii): $	heta=	heta_{2}=180^{\circ}$

$	herefore$ Torque, $	au=M B \sin 	heta$

$=M B \sin 180^{\circ}=0 \mathrm{~J}$

Question 8

Q. 5.8. A closely wound solenoid of 2000 turns and area of cross-section $1.6 	imes 10^{-4} \mathrm{~m}^{2}$, carrying a current of 4.0 A , is suspended through its centre allowing it to turn in a horizontal plane.

(a) What is the magnetic moment associated with the solenoid?

(b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of $7.5 	imes 10^{-2} \mathrm{~

Accepted Answer

Number of turns on the solenoid, $n=2000$

Area of cross-section of the solenoid, $\mathrm{A}=1.6 	imes 10^{-4} \mathrm{~m}^{2}$

Current in the solenoid, $\mathrm{I}=4 \mathrm{~A}$

(a)The magnetic moment along the axis of the solenoid is calculated as:

$M=n A I$

$=2000 	imes 1.6 	imes 10^{-4} 	imes 4$

$=1.28 \mathrm{Am}^{2}$

(b) Magnetic field, $B=7.5 	imes 10^{-2} T$

Angle between the magnetic field and the axis of the solenoid, $	heta=30^{\circ}$

Torque, $	au=M B \sin 	heta$

$=1.28 	imes 7.5 	imes 10^{-2} \sin 30^{\circ}$

$=4.8 	imes 10^{-2} \mathrm{Nm}$

Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is $4.8 	imes 10^{-2} \mathrm{Nm}$.

5. Magnetism and Matter

Exercises - 1