Magnetic Dipole

MAGNETIC DIPOLE Pole Strength Magnetic Dipole and Magnetic Dipole Moment Every magnet possesses two poles, ‘N’ and ‘S’. Similar poles of two magnets repel each other, while opposite poles attract each other, forming an action-reaction pair. [Image showing two bar magnets. The first diagram shows the North pole of one magnet facing the North pole of another, with arrows indicating repulsion. The second diagram shows the North pole of one magnet facing the South pole of another, with arrows indicating attraction.] The poles of a single magnet do not converge due to mutual attraction, and efforts to obtain two isolated poles by breaking the magnet in the middle are futile because the other end becomes a pole of opposite nature. As a result, ‘N’ and ‘S’ poles persist together. [Image showing a bar magnet being broken into two pieces, illustrating that each new piece still has a North and a South pole.] These poles are identified as positive and negative poles. The North Pole is considered the positive pole, equivalent to positive magnetic charge, while the South Pole is regarded as the negative pole, akin to negative magnetic charge, obeying the rules of convention. This is called “POLE STRENGTH,” denoted as +m for the North Pole and –m for the South Pole, analogous to the charges +q and –q in electrostatics. Pole strength is a scalar quantity and signifies the potency of the pole and, consequently, of the magnet. A magnet can be viewed as a dipole, as it invariably possesses two opposing poles (similar to an electric dipole with opposite charges +q and +q). This is referred to as a MAGNETIC DIPOLE, and it possesses a MAGNETIC DIPOLE MOMENT, symbolized by M. It is a vector quantity. Its direction is from –m to +m that means from S to N. [Image of a bar magnet with its poles labeled N and S, and pole strengths +m and -m. The magnetic length l_m and geometric length l_g are indicated.] M = m ⋅ l_m, here l_m => magnetic length of the magnet. l_m is slightly less than l_g it is the geometric length of the magnet. The poles (+m and –m) are located precisely at the ends of the magnet. For computational purposes, we can assume l_m ≈ l_g (Actually l_m/l_g ≈ 0.84). The units for ‘m’ and ‘M’ will be discussed later for your recollection and comprehension. Magnetic Field and Strength of Magnetic Field. The region surrounding a magnetic pole has a unique impact, causing the other pole to undergo a force. This distinctive influence is termed the MAGNETIC FIELD, and the resulting force is referred to as MAGNETIC FORCE. The qualitative representation of this field is conveyed through the STRENGTH OF MAGNETIC FIELD, also known as MAGNETIC INDUCTION or MAGNETIC FLUX DENSITY,” symbolized by B. It is essential to note that the magnetic field strength is a vector quantity. Definition of B The force encountered by a north pole with unit pole strength at a specific location, arising from other poles (referred to as the source), is termed the magnetic field strength at that point due to the source. B = F/m Here F = Magnetic force on pole of pole strength m. m may be +ve or –ve and of any value. S.I. unit of B is Tesla or Weber /m² (abbreviated as T and Wb/m²). We can also write F = mB. According to this direction of on +ve pole (North pole) will be in the direction of field and on –ve pole (south pole) it will be opposite to the direction of B [Image showing a North pole and a South pole with arrows indicating the direction of the magnetic field B and the force F.] The field produced by sources is independent of the test pole, regardless of its magnitude or sign. 1. B Due to various source (a) Due to a single pole: (Similar to the case of a point charge in electrostatics) B = (μ₀/4π) * (m/r²) This is magnitude The directions of B caused by the North pole and the South Pole are depicted as follows: [Two diagrams showing the direction of the magnetic field lines (B) originating from a North pole (N) and terminating at a South pole (S).] In vector form B = (μ₀/4π) * (m/r³) * r Here m is with sign and r = position vector of the test point with respect to the pole. (b) Due to a bar magnet: (Same as the case of electric dipole in electrostatics) Independent case never found. Always ‘N’ and ‘S’ exist together as magnet. [Diagram of a bar magnet with poles S and N, showing the magnetic field at a point A on the axis and a point B on the equatorial line.] At A (on the axis) = 2(μ₀/4π) * (M/r³) for a << r At B (on the equatorial) = -(μ₀/4π) * (M/r³) for a << r At General point B_a = 2(μ₀/4π) * (Mcosθ/r³) B_e = (μ₀/4π) * (Msinθ/r³) B_net = (μ₀M/4πr³) * √(1 + 3cos²θ) tan α = (B_e / B_a) = (tanθ / 2) [Diagram showing the vector components of the magnetic field (B_a, B_e, B_net) at a general point relative to a magnetic dipole.] Magnetic Lines of Force of a Bar Magnet [Diagram showing the magnetic field lines around a bar magnet, originating from the North pole and entering the South pole.] Ex. Find the magnetic force on a short magnet of magnetic dipole moment M₂ due to another short magnet of magnetic dipole M₁. [Diagram showing two co-axial short bar magnets.] Sol. To find the magnetic force F_res we must find the B field due to one magnet. We will also assume m and –m as pole strengths of ‘N’ and ‘S’ of M₂. Also length of M₂ as 2a. B₁ and B₂ are the strengths of the magnetic field due to M₁ at +m and –m respectively. They experience magnetic forces F₁ and F₂ as shown. F₁ = (μ₀/4π) * (2M₁/(r-a)³) m F₂ = (μ₀/4π) * (2M₁/(r+a)³) m F_res = F₁ - F₂ = 2(μ₀/4π) * (M₁m) * [1/(r-a)³ - 1/(r+a)³] = 2(μ₀/4π) * (M₁m/r³) * [(1 - a/r)⁻³ - (1 + a/r)⁻³] Through the utilization of acceleration, binomial expansion, and neglecting terms of high power, we obtain. F_res = 2(μ₀/4π) * (M₁m/r³) * [1 - (-3a/r) - {1 + (-3a/r)}] = 2(μ₀/4π) * (M₁m/r³) * (6a/r) = (μ₀/4π) * (6M₁M₂/r⁴) = G(M₁M₂/r⁴) Direction of F_res is towards right. Alternative Method B = (μ₀/4π) * (2M₁/r³) => dB/dr = (μ₀/4π) * (2M₁ * -3/r⁴) P = -M₂ * (dB/dr) => F = (μ₀/4π) * (6M₁M₂/r⁴) Ex. A magnet is 10 cm long and its pole strength is 120 CGS units (1 CGS unit of pole strength = 0.1 A-m). Find the magnitude of the magnetic field B at a point on its axis at a distance 20 cm from it. Sol. The pole strength is m = 120 CGS units = 12A-m. Magnetic length is 2a = 10 cm or l = 0.05 m. Distance from the magnet is d = 20 cm = 0.2 m. The field B at a point in end-on position is. B = (μ₀/4π) * (2Md / (d² - l²)²) = (10⁻⁷ * 2 * (12*0.1) * (0.2)) / ((0.2)² - (0.05)²)² T = 3.4 x 10⁻³ T Ex. Find the magnetic field due to a dipole of magnetic moment 1.2 A-m² at a point 1 m away from it in a direction making an angle of 60° with the dipole-axis. Sol. The magnitude of the field is B = (μ₀/4π) * (M/r³) * √(1 + 3cos²θ) = (10⁻⁷ * (1.2/1³)) * √(1 + 3cos²60°) = 1.6 x 10⁻⁷ T The direction of the field makes an angle α with the radial line where tan α = tanθ/2 = tan60°/2 Magnet in an External Uniform Magnetic Field (Same as case of electric dipole) F_res = 0 (for any angle) τ = M B sin θ Here θ is angle between M and B. [Diagram showing a magnetic dipole (m, -m) in a uniform magnetic field B, experiencing a torque τ.] Note τ acts such that it tries to make M x B. τ is same about every point of the dipole it’s potential energy is θ = 0° is stable equilibrium U = -M B cos θ = -M ⋅ B θ = π is unstable equilibrium For small ‘θ’ the dipole performs SHM about θ = 0° position τ = -M B sin θ; I α = -M B sin θ For small θ, sin θ ≈ θ α = -(MB/I) θ [Diagram showing the alignment of a magnetic dipole M in a field B.] Angular frequency of SHM ω = √(MB/I) T = 2π/ω ⇒ T = 2π√(I/MB) Here I = I_cm if the dipole is free to rotate = I_hinge if the dipole is hinged Ex. A bar magnet having a magnetic moment of 1.0 x 10⁻⁴ J/T is free to rotate in a horizontal plane. A horizontal magnetic field B = 4 x 10⁻⁵ T exists in the space. Find the work done in rotating the magnet slowly from a direction parallel to the field to a direction 60° from the field. Sol. The work done by the external agent is W = U_f - U_i = -MBcosθ_f - (-MBcosθ_i) = -MB(cos60° - cos0°) = -(1.0 x 10⁻⁴)(4 x 10⁻⁵)(1/2 - 1) = 0.2J Ex. A bar magnet of mass 100 g, length 7.0 cm, width 1.0 cm and height 0.50 cm takes π/2 seconds to complete an oscillation in an oscillation magnetometer placed in a horizontal magnetic field of 25μT. (a) Find the magnetic moment of the magnet. (b) If the magnet is put in the magnetometer with its 0.50 cm edge horizontal, what would be its time period. Sol. (a) The moment of inertia of the magnet about the axis of rotation is I = m/12 (l² + b²) = (100x10⁻³)/12 ((7x10⁻²)² + (1x10⁻²)²) [kg-m²] = 4.2 x 10⁻⁵ Kg-m² We have T = 2π√(I/MB) or, M = (4π²I)/(T²B) = (4π² * 4.2x10⁻⁵)/( (π/2)² * 25x10⁻⁶) = 27A - m². (b) In this case the moment of inertia becomes I' = m/12 (l² + b'²) where b' = 0.5 cm. The time period would be T' = 2π√(I'/MB) ... (2) Dividing by equation (1), T'/T = √(I'/I) = √((l²+b'²)/(l²+b²)) = √( ((7cm)²+(0.5cm)²) / ((7cm)²+(1.0cm)²) ) = 0.992 or, T' = 0.992T/2 s = 0.496πs Magnet in an External Non-uniform Magnetic Field In such problems, no specific formulas are employed. Instead, examine the force on individual poles and compute the resultant torque on the dipole. Ex. Find the torque on M₁ due to M₂ in Que. 1 Sol. Let B₁, B₂ be magnetic field of M₂ at ‘m’ of M₁ are B₁ and B₂ respectively. The forces on –m and +m are F₁ and F₂ as shown in the figure. The torque (about the centre of the dipole m₁) will be F₁a + F₂a = (F₁ + F₂)a = (B₁ + B₂) m + B_c m(-2a) = (B₁ + B₂) m + (B₁ + B₂)/2 (-2am) = m(B₁ + B₂) (a-a) V a < r = (μ₀/4π) * (2M₂/r³) * (2a/r) * m = (μ₀/4π) * (2M₁M₂/r³) [Diagram showing two perpendicular bar magnets and the forces involved.] Terrestrial Magnetism (Earth’s Magnetism) The notion that Earth possesses magnetism was initially proposed in the late sixteenth century by Dr. William Gilbert. Although the origin of Earth’s magnetism remains a subject of speculation among scientists, there is consensus that Earth acts as a magnetic dipole with a slight inclination (11.5°) to its axis of rotation, with the south pole pointing north. The figure illustrates the lines of force in Earth’s magnetic field, which run parallel to the Earth’s surface at the equator and perpendicular to it near the poles. When delving into the Earth’s magnetism, it is important to bear in mind that: [Diagram of the Earth showing its geographic and magnetic axes, with the magnetic field lines illustrated.] At a specific location, the magnetic meridian is not a mere line but rather a vertical plane that traverses through the axis of a freely suspended magnet. Essentially, it is a plane encompassing both the location and the. The geographical meridian of a place is a vertical plane that intersects the line connecting the geographical north and south. In essence, it is a plane encompassing both the location and the Earth’s axis of rotation. This plane passes through the geographical. The magnetic Equator is a substantial circle, centered at the Earth’s core, positioned on the Earth’s surface and perpendicular to the magnetic axis. The magnetic equator, passing through Trivandrum in South India, divides the Earth into two hemispheres.