Physics INTRODUCTION, THE BAR MAGNET FOR CBSE -NCERT 1

### Topic covered

color{blue}{star} INTRODUCTION
color{blue}{star} THE BAR MAGNET

### INTRODUCTION

color{blue} ✍️ Some of the commonly known ideas regarding magnetism are :

color{blue}{(i)} The earth behaves as a magnet with the magnetic field pointing approximately from the geographic south to the north.

color{blue} {(ii)} When a bar magnet is freely suspended, it points in the north-south direction. The tip which points to the geographic north is called the north pole and the tip which points to the geographic south is called the south pole of the magnet.

color{blue}{(iii)} There is a repulsive force when north poles ( or south poles ) of two magnets are brought close together. Conversely, there is an attractive force between the north pole of one magnet and the south pole of the other.

color{blue} {(iv) } We cannot isolate the north, or south pole of a magnet. If a bar magnet is broken into two halves, we get two similar bar magnets with somewhat weaker properties. Unlike electric charges, isolated magnetic north and south poles known as magnetic monopoles do not exist.

color{blue} {(v)} It is possible to make magnets out of iron and its alloys. We begin with a description of a bar magnet and its behaviour in an external magnetic field. We describe Gauss’s law of magnetism. We then follow it up with an account of the earth’s magnetic field. We next describe how materials can be classified on the basis of their magnetic properties. We describe para-, dia-, and ferromagnetism. We conclude with a section on electromagnets and permanent magnets.

### THE BAR MAGNET

color{blue} ✍️ The pattern of iron filings suggests that the magnet has two poles similar to the positive and negative charge of an electric dipole. As mentioned in the introductory section, one pole is designated the North pole and the other, the South pole. When suspended freely, these poles point approximately towards the geographic north and south poles, respectively. A similar pattern of iron filings is observed around a current carrying solenoid. The arrangement of iron filings is shown in Fig. 5.2.

color{green}bbul("The magnetic field lines")
color {blue}{➢➢}The pattern of iron filings permits us to plot the magnetic field lines*. This is shown both for the bar-magnet and the current-carrying solenoid in Fig. 5.3. For comparison refer to the Chapter 1, Figure 1.17(d). Electric field lines of an electric dipole are also displayed in Fig. 5.3(c). The magnetic field lines are a visual and intuitive realisation of the magnetic field. Their properties are:

color{blue} {(i)} The magnetic field lines of a magnet (or a solenoid) form continuous closed loops. This is unlike the electric dipole where these field lines begin from a positive charge and end on the negative charge or escape to infinity.

color{blue} {(ii)} The tangent to the field line at a given point represents the direction of the net magnetic field color{blue}(B) at that point.

color{blue}{(iii)} The larger the number of field lines crossing per unit area, the stronger is the magnitude of the magnetic field color{blue}(B). In Fig. 5.3(a), B is larger around region ii than in region i .

color{blue} {(iv) } The magnetic field lines do not intersect, for if they did, the direction of the magnetic field would not be unique at the point of intersection. One can plot the magnetic field lines in a variety of ways. One way is to place a small magnetic compass needle at various positions and note its orientation. This gives us an idea of the magnetic field direction at various points in space.

Q 3154080854

In Fig. 5.4(b), the magnetic needle has magnetic moment
6.7 × 10–2 Am2 and moment of inertia  = 7.5 × 10–6 kg m^2. It performs
10 complete oscillations in 6.70 s. What is the magnitude of the
magnetic field?
Class 12 Chapter example 1
Solution:

The time period of oscillation is,

T = (6.70)/(10)= 0.67s

From Eq. (5.5)
B = (4pi^2g)/(mT)

= (4xx(3.14)^2xx7.5xx10^(-6))/(6.7 xx10^(-2) xx(0.67)^2

= 0.001T
Q 2637001882

In Fig. the magnetic needle has magnetic moment 6.7 xx 10^(–2) Am^2 and moment of inertia g = 7.5 × 10^(–6) kg m^2. It performs
10 complete oscillations in 6.70 s. What is the magnitude of the magnetic field?

Solution:

The time period of oscillation is,

T = (6.70)/(10) = 0.67 s

we have B = (4 pi^2 g)/(m T^2 )

 = (4xx (3.14)^2 xx 7.5 xx 10^(-6))/(6.7xx10^(-2) xx (0.67)^2)

 = 0.01 T
Q 2647101983

What is the magnitude of the equatorial and axial fields due to a bar magnet of length 5.0 cm at a distance of 50 cm from its
mid-point? The magnetic moment of the bar magnet is 0.40 A m^2,

Solution:

We have B_E = (mu_0 m)/(4 pi r^3 ) = (10^(-7) xx 0.4)/(0.5)^3 = (10^(-7) xx 0.4)/(0.125) = 3.2xx10^(-7) T

B_A = (mu_0 2 m)/(4 pi r^3) = 6.4xx10^(-7) T
Q 3174180956

Figure 5.5 shows a small magnetised needle P placed at a point O. The arrow shows the direction of its magnetic moment. The other arrows show different positions (and orientations of the magnetic moment) of another identical magnetised needle Q.
(a) In which configuration the system is not in equilibrium?
(b) In which configuration is the system in (i) stable, and (ii) unstable equilibrium?
(c) Which configuration corresponds to the lowest potential energy among all the configurations shown?
Class 12 Chapter example 5
Solution:

Potential energy of the configuration arises due to the potential energy of
one dipole (say, Q) in the magnetic field due to other (P). Use the result
that the field due to P is given by the expression [Eqs. (5.7) and (5.8)]:

B_P - (mu_0)/(4pi) (m_P)/(r^3) (on the normal bisector)

B_P=(mu_0 \ \ 2)/(4pi) (m_P)/(r^3) (on the axis)
where m_P is the magnetic moment of the dipole P.
Equilibrium is stable when m_Q is parallel to B_P,  and unstable when it is anti-parallel to B_P.
For instance for the configuration Q_3 for which Q is along the perpendicular bisector of the dipole P, the magnetic moment of Q is parallel to the magnetic field at the position 3. Hence Q_3 is stable.
Thus,
"(a)" \ \ PQ_1 and PQ_2
"(b)" \ \ (i) PQ_3, PQ_6 "(stable);" (ii) PQ_5, PQ_4 "(unstable)"
"(c)" \ \ PQ_6

### Bar magnet as an equivalent solenoid

color{blue} ✍️ In the previous chapter, we have explained how a current loop acts as a magnetic dipole (Section 4.10). We mentioned Ampere’s hypothesis that all magnetic phenomena can be explained in terms of circulating currents.

color {blue}{➢➢} Recall that the magnetic dipole moment color{blue}(m) associated with a current loop was defined to be color{blue}(m = NI A) where color{blue}(N) is the number of turns in the loop, color{blue}(I) the current and color{blue}(A) the area vector (Eq. 4.30).

color{blue} ✍️ The resemblance of magnetic field lines for a bar magnet and a solenoid suggest that a bar magnet may be thought of as a large number of circulating currents in analogy with a solenoid.

color {blue}{➢➢} Cutting a bar magnet in half is like cutting a solenoid. 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.

color {blue}{➢➢} One can test this analogy by moving a small compass needle in the neighbourhood of a bar magnet and a current-carrying finite solenoid and noting that the deflections of the needle are similar in both cases.

color {blue}{➢➢} To make this analogy more firm we calculate the axial field of a finite solenoid depicted in Fig. 5.4 (a). We shall demonstrate that at large distances this axial field resembles that of a bar magnet.

color{blue} ✍️ Let the solenoid of Fig. 5.4(a) consists of color{blue}(n) turns per unit length. Let its length be color{blue}(2l) and radius a. We can evaluate the axial field at a point color{blue}(P), at a distance color{blue}(r) from the centre color{blue}(O) of the solenoid. To do this, consider a circular element of thickness color{blue}(dx) of the solenoid at a distance color{blue}(x) from its centre. It consists of color{blue}(n d x) turns.

color{blue} ✍️ Let color{blue}(I) be the current in the solenoid. In Section 4.6 of the previous chapter we have calculated the magnetic field on the axis of a circular current loop. From Eq. (4.13), the magnitude of the field at point color{blue}(P) due to the circular element is

color{blue}(dB= (mu_0ndxIa^2)/(2[(r-x)^2+a^2]^(3//2)))

color {blue}{➢➢} The magnitude of the total field is obtained by summing over all the elements — in other words by integrating from color{blue}(x = – l) to color{blue}(x = + l). Thus,

color{blue}(B= (mu_0 nla^2)/2 int_(-l)^(l) (dx)/([(r-x)^2+a^2]^(3//2)))

color {blue}{➢➢} This integration can be done by trigonometric substitutions. This exercise, however, is not necessary for our purpose. Note that the range of color{blue}(x) is from color{blue}(– l) to color{blue}(+ l). Consider the far axial field of the solenoid, i.e., color{blue}(r > > a) and color{blue}(r > > l) . Then the denominator is approximated by

color{blue}([(r-x)^2+a^2]^(3//2) approx r^3)

color {blue}{➢➢} and color{bue}(B = (mu_0 nIa^2)/(2r^3) int_(-l)^(l) dx)

color{blue}(= (mu_0 nI)/(2) (2la^2)/(r^3))

...................(5.1)

color {brown}bbul{"Note"} that the magnitude of the magnetic moment of the solenoid is, color{blue}(m = n (2l) I (πa^2)) — (total number of turns × current × cross-sectional area). Thus,

color{blue}(B= (mu_0)/(4pi) (2m)/(r^3))

...............(5.2)

color {blue}{➢➢} This is also the far axial magnetic field of a bar magnet which one may obtain experimentally. Thus, a bar magnet and a solenoid produce similar magnetic fields.

color {blue}{➢➢} The magnetic moment of a bar magnet is thus equal to the magnetic moment of an equivalent solenoid that produces the same magnetic field.

color{blue} ✍️A magnetic charge (also called pole strength) color{blue}(+q_m) to the north pole and color{blue}(–q_m) to the south pole of a bar magnet of length color{blue}(2l), and magnetic moment color{blue}(q_m(2l )).

color {blue}{➢➢} The field strength due to color{blue}(q_m) at a distance r from it is given by color{blue}(μ_0q_m//4πr^2). The magnetic field due to the bar magnet is then obtained, both for the axial and the equatorial case, in a manner analogous to that of an electric dipole.

color {blue}{➢➢} The method is simple and appealing. However, magnetic monopoles do not exist, and we have avoided this approach for that reason.

### The dipole in a uniform magnetic field

color{blue} ✍️ The pattern of iron filings, i.e., the magnetic field lines gives us an approximate idea of the magnetic field color{blue}(B). We may at times be required to determine the magnitude of color{blue}(B) accurately.

color {blue}{➢➢}This is done by placing a small compass needle of known magnetic moment color{blue}(m) and moment of inertia color{blue}(I) and allowing it to oscillate in the magnetic field. This arrangement is shown in Fig. 5.4(b).

color {blue}{➢➢} The torque on the needle is [see Eq. (4.29)],

color{blue}(τ = m × B)

.......................(5.3)

color {blue}{➢➢} In magnitude color{blue}(τ = mB sinθ)

color {blue}{➢➢} Here color{blue}(τ) is restoring torque and color{blue}(θ) is the angle between color{blue}(m) and color{blue}(B).

color {blue}{➢➢} Therefore, in equilibrium color{blue}(I (d^2theta)/(dt^2) = - mB sin theta)

color {blue}{➢➢} Negative sign with color{blue}(mB sinθ) implies that restoring torque is in opposition to deflecting torque. For small values of color{blue}(θ) in radians, we approximate sin color{blue}(θ ≈ θ) and get

color{blue}(I (d^2theta)/(dt^2) = - mB theta)

color{blue}((d^2theta)/(dt^2) =-(mB)/I theta)

color {blue}{➢➢}This represents a simple harmonic motion. The square of the angular frequency is color{blue}(ω^2 = mB//I) and the time period is,

color{blue}(T = 2pisqrt(I/(mB))

.......................(5.4)

or

color{blue}(B = (4pi^2I)/(mT^2))

.......................(5.5)

color {blue}{➢➢} An expression for magnetic potential energy can also be obtained on lines similar to electrostatic potential energy. The magnetic potential energy color{blue}(U_m) is given by

color{blue}(U_m = int tau(theta)d theta)

color{blue}(= int mB sin theta = - Mb cos theta)

color{blue}(= −m * B)

.............................(5.6)

color {blue}{➢➢} We have emphasised in Chapter 2 that the zero of potential energy can be fixed at one’s convenience. Taking the constant of integration to be zero means fixing the zero of potential energy at color{blue}(θ = 90^º), i.e., when the needle is perpendicular to the field. Equation (5.6) shows that potential energy is minimum color{blue}((= –mB)) at color{blue}(θ = 0^º) (most stable position) and maximum color{blue}((= +mB)) at color{blue}(θ = 180^º) (most unstable position).

### The electrostatic analog

color {blue}{➢➢} Comparison of Eqs. (5.2), (5.3) and (5.6) with the corresponding equations for electric dipole (Chapter 1), suggests that magnetic field at large distances due to a bar magnet of magnetic moment m can be obtained from the equation for electric field due to an electric dipole of dipole moment p, by making the following replacements:

color{blue}(E →B , p → m 1/(4piε_0) -> (mu_0)/(4pi))

color {blue}{➢➢} In particular, we can write down the equatorial field color{blue}(B_E) of a bar magnet at a distance color{blue}(r), for color{blue}(r > > l), where color{blue}(l) is the size of the magnet

color{blue}(B_E = - (mu_0m)/(4pir^3))

............(5.7)

color {blue}{➢➢} Likewise, the axial field color{blue}(B_A) of a bar magnet for color{blue}(r >> l) is:

color{blue}(B_A = (mu_0)/(4pi) (2m)/(r^3))

.............(5.8)

color {blue}{➢➢} Equation (5.8) is just Eq. (5.2) in the vector form. Table 5.1 summarises the analogy between electric and magnetic dipoles.