Contents
Physics Topics can help us understand the behavior of the natural world around us.
What is the Formula for the Magnetic Field of a Bar Magnet?
End-on or axial position : A bar magnet SN is kept in air [Fig.]. Its pole-strength is qm and magnetic length = SN = 2l. A point P is taken on the axis of the magnet at a distance d from its mid-point O. The position of the point P is called end-on or axial position. We have to determine the magnetic field at the point P due to the magnet.
Calculation: Magnetic field at the point P due to N -pole,
B1 = \(\frac{\mu_0}{4 \pi} \cdot \frac{q_m}{(d-l)^2}\) ; the direction of \(\vec{B}_1\) is along \(\overrightarrow{O P}\)
Again, magnetic field at the point P due to S -pole,
B2 = \(\frac{\mu_0}{4 \pi} \cdot \frac{q_m}{(d+l)^2}\) ; the direction of \(\overrightarrow{B_2}\) is along \(\overrightarrow{P O}\)
∵ N-pole is nearer to P than S-pole, the value of B1 is greater than that of B2.
Therefore, the resultant magnetic field at the point P,
The direction of \(\vec{B}\) is along \(\overrightarrow{O P}\).
The vector representation of the equation is,
\(\vec{B}\) = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 \vec{p}_m d}{\left(d^2-l^2\right)^2}\)
The resultant magnetic intensity in CGS system,
H = \(\frac{2 p_m d}{\left(d^2-l^2\right)^2}\)
If the length of the bar magnet is very small and the point P is taken at a very large distance from the bar magnet, i.e., d\(\gg\)1, then (d2 – l2) ≈ d4.
∴ B = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 p_m}{d^3}\)
Broadside-on or equatorial position: A bar magnet SN is kept in air (Fig. 2.111. Its pole-strength is and magnetic length = SN = 2l. A point P is taken on the perpendicular bisector of the magnetic-length and at a distance d from the midpoint O of the magnet. The position of the point P is called broadside-on or equatorial position. We have to determine the magnetic field at the point P due to the magnet.
Calculation: Magnetic field at the point P due to N-pole,
B1 = [lartex]\frac{\mu_0}{4 \pi} \cdot \frac{q_m}{N P^2}[/latex]; the direction of [lartex]\vec{B}_1[/latex] is along [lartex]\overrightarrow{P Q}[/latex]
Again, magnetic field at the point P due to S -pole,
B2 = \(\frac{\mu_0}{4 \pi} \cdot \frac{q_m}{S P^2}\); the direction of \(\vec{B}_2\) is along \(\overrightarrow{P S}\)
The components of B1 and B2, i.e., B1sinθ and B2 sinO, along PT and PO respectively, cancel each other.
On the other hand, the components B1 cosθ and B2 cosθ along PR are added together.
So, the resultant magnetic field at the point P,
B = 2B1 cosθ [∵ magnitudes of B1, B2 are equal]
or B = \(\frac{2 \mu_0}{4 \pi} \cdot \frac{q_m}{N P^2} \cos \theta\) = \(\frac{2 \mu_0}{4 \pi} \cdot \frac{q_m}{\left(d^2+l^2\right)} \cdot \frac{l}{\sqrt{d^2+l^2}}\)
= \(\frac{\mu_0}{4 \pi} \cdot \frac{p_m}{\left(d^2+l^2\right)^{3 / 2}}\) ; the direction of \(\vec{B}\) along \(\overrightarrow{P R}\)
The vector representation of the equation is,
\(\vec{B}\) = \(-\frac{\mu_0}{4 \pi} \cdot \frac{\vec{p}_m}{\left(d^2+l^2\right)^{3 / 2}}\)
The resultant magnetic intensity in CGS system,
H = \(\frac{p_m}{\left(d^2+l^2\right)^{3 / 2}}\)
If the length of the bar magnet is very small and the point P is taken at a very large distance from the bar magnet, i.e., l\(\ll\)d
then, B = \(\frac{\mu_0}{4 \pi} \cdot \frac{p_m}{d^3}\)
So, magnetic field at the end-on or axial position
= 2 × magnetic field at the broadside-on or equatorial position
Any position: With respect to the bar magnet of magnetic length 2l, the point P under consideration is at the position (r, θ), i.e., the point P is at a distance r from the mid-point of the magnet and lies at an angle θ with the magnetic axis [Fig.].
The component of pm along r is pmcosθ; with respect to this component, the point P is at the end- on position and hence the magnetic field at the point Br = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 p_m \cos \theta}{r^3}\) (∵ r \(\gg\) l)
Again, the component of pm normal to r is pmsinθ; with respect to this component the point P is at the equatorial position, and hence the magnetic field at the point P,
Bθ = \(\frac{\mu_0}{4 \pi} \cdot \frac{p_m \sin \theta}{r^3}\)
So, according to Fig., the resultant magnetic field at the point P,
Bθ = \(\sqrt{B_r^2+B_\theta^2}\) = \(\sqrt{\left(\frac{\mu_0}{4 \pi} \cdot \frac{p_m}{r^3}\right)^2\left(4 \cos ^2 \theta+\sin ^2 \theta\right)}\)
= \(\frac{\mu_0}{4 \pi} \cdot \frac{p_m}{r^3} \sqrt{3 \cos ^2 \theta+1}\)
The direction of \(\vec{B}\) is along \(\overrightarrow{P R}\); where ∠TPR = ϕ (say)
∴ tan ϕ = \(\frac{B_\theta}{B_r}\) = \(\frac{1}{2} \tan \theta\)
The resultant magnetic intensity in CGS system,
H = \(\frac{p_m}{r^3} \sqrt{3 \cos ^2 \theta+1}\)
Equivalence of a solenoid and a bar magnet: Let us assume a solenoid of radius = a, length = 2l, number of turns per unit length = n. Current through it = I. So, the total number of turns of the solenoid = 2ln and hence the effective net cur-rent around its axis = 2lnI. As the cross sectional area of the solenoid = πa2, so, the magnetic dipole moment of the current carrying solenoid, pm = (2lnl) ᐧ (πa2) = 2πlna2I.
Now the centre (O) of the solenoid is taken as origin and its axis as x -axis [Fig.]. Considering a small length dx at a distance x from O, we get a circular current carrying coil with number of turns ndx. For this coil, magnetic field thus produced at a point P on the axis (where, OP = r and r\(\gg\), r\(\gg\)x),
dB = \(\frac{\mu_0(n d x) I}{2} \cdot \frac{a^2}{\left\{a^2+(r-x)^2\right\}^{3 / 2}}\) = \(\frac{\mu_0 n I a^2}{2 r^3} d x\)
[As a and x are negligible with respect to r, so {a2 + (r – x)2}3/2 ≈ r3]
So, the magnetic field at P for the entire solenoid,
B = \(\frac{\mu_0 n I a^2}{2 r^3} \int_{-l}^l d x\) = \(\frac{\mu_0 n I a^2}{2 r^3} \cdot 2 l\) = \(\frac{\mu_0}{4 \pi} \cdot \frac{2\left(2 \pi \ln a^2 I\right)}{r^3}\)
i.e., B = \(\frac{\mu_0}{4 \pi} \frac{2 p_m}{r^3}\)
In this very section we have already found the same expression due to a bar magnet [see end on or axial position]. Thus we con-clude that a bar magnet and a current carrying solenoid are equivalent to each other.
Comparing their magnetic moments, we get,
qm ᐧ 2l = 2πlna2I or, qm = πna2l
Magnetic Moment of Charged Particle Moving in a Circle
We know that, a revolving charged particle is equivalent to an electric current. So, the orbit of that particle is equivalent to a current loop and as a result, it behaves as a magnetic dipole which must have a magnetic moment.
Suppose a particle having a charge q is revolving in a plane circular path; r = radius of the orbit of that particle and v = velocity of the particle [Fig.].
Time period of revolution of the particle, T = \(\frac{2 \pi r}{v}\)
Equivalent current, I = \(\frac{q}{T}\) = \(\frac{q v}{2 \pi r}\)
So, the magnetic moment of the particle,
pm = \(\frac{q v}{2 \pi r} \cdot A\) = \(\frac{q v}{2 \pi r} \cdot \pi r^2\) = \(\frac{q v r}{2}\)
The direction of this magnetic moment pm can be obtained by applying corkscrew rule. If the mass of the particle be m then along the axis of the circular path, i.e., along the direction of pm, the angular momentum of the particle,
L = mvr [L = moment of momentum
= momentum × radius of circular path]
∴ pm = \(\frac{q v r}{2}\) = \(\frac{q}{2 m} \cdot m v r\) = \(\frac{q}{2 m} L\)
Since the directions of pm and L are the same, using vector notation we can write,
\(\vec{p}_m\) = \(\frac{q}{2 m} \vec{L}\) ……. (1)
Magnetic dipole moment of an electron: Electrons inside an atom revolve around the nucleus continuously. As a revolving charged particle, each electron behaves as a magnetic dipole. Substituting the charge of an electron -e in place of q in equation (1), we can write,
\(\vec{p}_m\) = \(-\frac{e}{2 m} \cdot \vec{L}\) ….. (2)
The negative sign on the right hand side indicates that, charge of an electron being negative, its angular momentum and magnetic moment are oppositely directed.
From Bohr’s theory related with atomic structure [see details in the chapter: Atom], it is known that electrons revolving in differ-ent orbits have discrete angular momenta, defined as
L = \(n \frac{h}{2 \pi}\) = nh [n = 1, 2, 3, ……..]
Here, h = Planck’s constant = 6.626 × 10-34 J ᐧ s and h = \(\frac{h}{2 \pi}\) = 1.05 × 10-34 J ᐧ s = reduced Planck’s constant.
So, the magnitude of magnetic moment of an electron [from equation (2)],
pm = \(\frac{e}{2 m} n \hbar\) = \(n \frac{e \hbar}{2 m}\) …….. (3)
In case of the innermost orbit of an atom (K-orbit), n = 1; then pm = \(\frac{e \hbar}{2 m}\) – this quantity is called Bohr magneton. It is denoted
by µB and it is the atomic unit of magnetic moment.
1µB = \(\frac{e \hbar}{2 m}\) = \(\frac{\left(1.6 \times 10^{-19} \mathrm{C}\right) \times\left(1.05 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)}{2 \times\left(9.1 \times 10^{-31} \mathrm{~kg}\right)}\)
= 9.23 × 10-24 A ᐧ m2
= 9.23 × 10-24 J ᐧ T-1
Spin of an electron: Besides the revolution of each electron in its orbit, it also rotates about its own axis (like diurnal motion of the earth). This is known as spin of an electron. An electron may have any of two oppositely directed spins. One kind of spin (say, clockwise) is called up spin and is denoted by ↑ symbol. The opposite kind of spin is then known as down spin and is denoted by the symbol ↓. Due to spin of any electron, a magnetic moment generates. By theoretical analysis its value is found to be 1 Bohr magneton. It is called the intrinsic spin magnetic moment.
The concept of rotation of an electron about its own axis is an oversimplified concept and requires quantum physics for proper explanation.
Magnetism of an atom: If the resultant moment i.e., vector sum of the magnetic moments of the electrons rotating around the nucleus inside an atom be zero, we can say that the atom as a whole does not show any magnetic property. On the other hand, if the resultant is not zero, the atom behaves like a magnetic dipole. Due to this property, each atom or molecule of iron, nickel etc. behaves like a small magnet and is called an atomic magnet.
Numerical Examples
Example 1.
A torque of 8 units is applied on a magnet when it is kept at 30° with the direction of a uniform magnetic field of intensity 0.32 units. Determine the magnetic moment of the magnet. [HS 06]
Solution:
If a magnet having magnetic moment pm is placed inclined at an angle θ with a uniform magnetic field B, the torque acting on the magnet,
\(\tau\) = pmB sinθ
Here, B = 0.32 units, θ = 30° and \(\tau\) = 8 units.
∴ pm = \(\frac{\tau}{B \sin \theta}\) = \(\frac{8}{0.32 \times \sin 30^{\circ}}\)
= \(\frac{8}{0.32 \times \frac{1}{2}}\) = 50 units
Example 2.
If the distance between two north poles of equal strength be 2 cm, the mutual force of repulsion between them becomes 2.5 dyn. What should be the distance of separation between them for which the repulsive force becomes 3.6 dyn ? [HS ’03]
Solution:
Magnetic force between two magnetic poles,
F = \(\frac{\mu_0}{4 \pi} \cdot \frac{q_m \cdot q_m^{\prime}}{r^2}\)
So, if the pole-strength remains unchanged then, F ∝ \(\frac{1}{r^2}\)
Hence, for two separate distances,
\(\frac{F_1}{F_2}\) = \(\left(\frac{r_2}{r_1}\right)^2\) or, \(\frac{r_2}{r_1}\) = \(\sqrt{\frac{F_1}{F_2}}\)
∴ r2 = \(r_1 \sqrt{\frac{F_1}{F_2}}\) = 2 × \(\sqrt{\frac{2.5}{3.6}}\) = 2 × \(\frac{5}{6}\) = 1.67 cm
Example 3.
The length of a bar magnet is 20 cm and its magnetic moment is 0.6 A m2. Determine the magnetic field at a point 30 cm away from either end.
Solution:
In Fig.,
PN = PS = 30 cm = 0.3 m = \(\sqrt{d^2+l^2}\)
Magnetic field at the point P due to the magnet,
B = \(\frac{\mu_0}{4 \pi} \cdot \frac{p_m}{\left(d^2+l^2\right)^{3 / 2}}\)
= \(\frac{4 \pi \times 10^{-7}}{4 \pi} \times \frac{0.6}{(0.3)^3}\) = 2.22 × 10-6Wb ᐧ m-2
Example 4.
The length of a bar magnet is 20 cm and its magnetic moment is 0.6 A ᐧ m2. Determine the magnetic field at a point on the axis of the magnet and 30 cm away from the north pole.
Solution:
Length of the magnet, 2l = 20 cm = 0.2 m
So, l = 0.1 m
Distance of the given point from the centre of the magnet,
d = (0.3 + 0.1)m = 0.4 m
∴ Magnetic field at the given point due to the magnet,
B = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 p_m d}{\left(d^2-l^2\right)^2}\) = \(\frac{4 \pi \times 10^{-7}}{4 \pi} \times \frac{2 \times 0.6 \times 0.4}{\left\{(0.4)^2-(0.1)^2\right\}^2}\) = \(\)
= 10-7 × \(\frac{0.48}{0.0225}\) = 2.13 × 10-6Wb ᐧ m-2
Example 5.
The radius of a circular conducting coil of 100 turns is 10 cm. If 2A current passes through the coil, what will be the magnetic moment generated?
Solution:
Magnetic moment,
pm = NIA = NI πr2
= 100 × 2 × (π × 0.1 × 0.1) [∵ 10 cm = 0.1m]
= 6.28 A ᐧ m2
Example 6.
If the magnetic moment of a straight magnetised wire is pm, what will be its magnetic moment when the wire is bent in the form of a semicircle? [WBJEE 2000]
Solution:
Let the length of the magnetised wire be l.
∴ Pole-strength of the wire, m = \(\frac{p_m}{l}\)
When the wire is bent in the form of semicircle, its effective length = diameter of the semicircle = 2r.
According to the problem, πr = l or, r = \(\frac{l}{\pi}\)
∴ The changed magnetic moment of the wire
= effective length × pole-strength
= 2r × \(\frac{p_m}{l}\) = \(\frac{2 l}{\pi} \cdot \frac{p_m}{l}\) = \(\frac{2 p_m}{\pi}\)
Example 7.
Magnetic moments of two small magnets are pm and pm; they are kept on a table as shown in the Fig. What will be the magnitude and direction of the mag-netic field produced by the magnets at the point P ?
[pm = 2.7 A m2,
P’m = 3.2 A m2;
d1 = 30 cm, d2 = 40 cm ]
Solution:
According to Fig., the point P lies on the perpen-dicular bisectors of both the magnets.
Magnetic field at the point P due to the magnet having magnetic moment pm,
B1 = \(\frac{\mu_0}{4 \pi} \cdot \frac{p_m}{d_1^3}\) = 10-7 × \(\frac{2.7}{(0.3)^3}\) = 10-5T
At the point , magnetic field due to the magnet having magnetic moment P’m is,
B2 = \(\frac{\mu_0}{4 \pi} \cdot \frac{p_m^{\prime}}{d_2^3}\) = 10-7 × \(\frac{3.2}{(0.4)^3}\) = 5 × 10-6T
Since, these two magnetic fields are perpendicular to each other, the resultant magnetic field at the point P,
B = \(\sqrt{B_1^2+B_2^2}\)
= \(\sqrt{\left(10^{-5}\right)^2+\left(5 \times 10^{-6}\right)^2}\)
= 1.12 × 10-5T
If the resultant magnetic field makes an angle θ with B1, then,
tanθ = \(\frac{B_2}{B_1}\) = \(\frac{5 \times 10^{-6}}{10^{-5}}\) = 0.5
∴ θ = tan-1(0.5) = 26.57°
Example 8.
“ Two bar magnets A and B, each having a magnetic length of 4 cm are placed along a straight line with their north poles 8 cm apart and facing each other. The neutral point lies on the axis, 2 cm from the north pole of the magnet A. Calculate the ratio of the magnetic moments of A and B. [Ignore the effect of earth’s magnetic field]. [HS’ 11]
Solution:
O is the neutral point [Fig.].
Given, NO = 2 cm
∴ N’O = (8 – 2) = 6 cm
XN = X’N’ = l = \(\frac{4}{2}\) cm = 2 × 10-2 m
d1 = XO = (2 + 2) cm = 4 × 10-3 m
d2 = X’O = (6 + 2) cm = 8 × 10-3 m
According to the question, B1 = B2
∴ \(\frac{\mu_0}{4 \pi} \cdot \frac{2 p_{m_1} \times 4 \times 10^{-2}}{144 \times 10^{-8}}\) = \(\frac{\mu_0}{4 \pi} \cdot \frac{2 p_{m_2} \times 8 \times 10^{-2}}{3600 \times 10^{-8}}\)
or \(\frac{p_{m_1}}{p_{m_2}}\) = \(\frac{2 \times 144}{3600}\) = \(\frac{2}{25}\)
∴ Ratio of the magnetic moments = \(p_{m_1}\) : \(p_{m_2}\) = 2 : 25