Magnetic Fields & Forces

• The document discusses the magnetic effects of current, providing derivations for various scenarios

Magnetic Field at the Center of a Circular Loop

• Considers a circular loop with current I to find the magnetic field at its center.
• Uses a small current element dl on the loop's circumference.
• The angle between dl and r (radius) is 90 degrees.
• Applying Biot-Savart's law yields an equation for dB (magnetic field due to dl).
• Integrating both sides determines the total magnetic field B.
• B = (µ₀I) / (2r), where µ₀ is the permeability of free space and r is the radius

Magnetic Field due to an Arc

• A complete circle can be treated as an arc subtending an angle of 2π.
• Magnetic field at the center of an arc can be found using the unitary method.
• For an angle of 2π, the magnetic field is (µ₀I) / (2r).
• For 1 radian, the magnetic field is (µ₀I) / (4πr).
• For any angle θ, the magnetic field B = (µ₀I / 4πr) * θ

Magnetic Field due to a Straight Conductor

• Focuses on finding the magnetic field at point P, a perpendicular distance r from a straight conductor
• The conductor carries current I.
• B = (µ₀I / 4πr) * (sin φ₁ + sin φ₂), where φ₁ and φ₂ are angles from point P to the ends of the conductor
• Considers special cases such as when the length of the wire is infinite or very long
• The value of r is very small
• If point P lies near one end, one angle is 90 degrees, and the other is 0, simplifying the equation to B = (µ₀I) / (4πr).
• If point P lies near the center, both angles approach 90 degrees, leading to B = (µ₀I) / (2πr).

Magnetic Field on the Axis of a Circular Loop

• Determines the small magnetic field due to a current element Idl of a circular loop.
• The loop has radius r and is at a point P at distance x from its center.
• dB = (µ₀ Idl) / (4π (r² + x²)).
• The dB cosΦ component from a current element is canceled by an opposite element.
• The dB sinΦ components add up to give the net magnetic field along the axis
• After integrating, B = (µ₀Ir²) / (2(r² + x²)^(3/2)).
• The direction is along the axis, towards the loop for clockwise current and away for anticlockwise.
• Special points: far from the center (x >> r), the field is B = (µ₀Ir²) / (2πx³) or B = (µ₀IA) / (2πx³), where A is the loop area.
• The magnetic field strength at the center is B = µ₀NI / 2r
• The magnetic field on the axis of a circular loop is B = (µ₀NIr²) / (2(r² + x²)^(3/2))

Ampere's Circuital Law

• The line integral of magnetic field intensity over a closed loop is µ₀ times the total current threading the loop.
• Mathematically, ∫ B ⋅ dl = µ₀I.
• The proof considers a straight conductor with current and a circular Amperian loop around it.
• B and dl are in the same direction, so the angle is 0.
• Applications include finding the magnetic field intensity at the center of a long solenoid.

Magnetic Field Intensity at the Centre of a Long Solenoid

• Considers a solenoid with n turns per unit length carrying current I
• Magnetic field lines inside are parallel to the axis; outside, the field is zero.
• Applying Ampere's law to a closed loop PQRS, we get ∫ B ⋅ dl = µ₀ × total current threading the loop PQRS.
• ∫ B ⋅ dl = ∫ B ⋅ dl (from P to Q) + ∫ B ⋅ dl (from Q to R) + ∫ B ⋅ dl (from R to S) + ∫ B ⋅ dl (from S to P)
• ∫ B ⋅ dl = BL as the only non-zero contribution.
• µ₀ × total current = µ₀nLI, leading to BL = µ₀nLI, or B = µ₀nI.
• At the ends of the solenoid, B = (1/2)µ₀nI.

Force Acting on a Charged Particle in a Magnetic Field

• A charge q moving with velocity v in a magnetic field B experiences a force F.
• F is proportional to q, v, B, and sin θ (angle between v and B).
• F = qvB sin θ.
• In vector form, F = q(v × B).
• If the charge enters perpendicularly, its path is circular.
• The centripetal force is provided by the magnetic force, leading to r = mv / qB.

Radius, Velocity, Time Period, and Frequency

• Radius of the circular path: r = mv / Bq
• Velocity: v = Bqr / m
• The time period T = 2πm / Bq
• The Frequency v = Bq / 2πm
• Angular frequency ω = Bq / m
• Kinetic energy KE = (1/2) m v² = (B²q²r²) / (2m)
• If the charge enters at an angle, its velocity splits into components causing helical motion.

Velocity Selector or Velocity Filter

• A charge moving perpendicularly to both electric and magnetic fields experiences forces that must be equal and opposite, qE = qvB
• The Velocity can then be defined as v = E / B.
• Used in velocity selectors or filters to select particles with a specific velocity.

Force Acting on a Current-Carrying Conductor

• Considers a conductor of length l and area A carrying current I in a magnetic field B at angle θ.
• Total number of electrons is Aln.
• Force on one electron is f = ev_d B sin θ, where v_d is the drift velocity.
• The total force is F = IlB sin θ.

Force Between Two Parallel Straight Conductors

• Considers two parallel conductors carrying current.
• Each creates a magnetic field affecting the other.
• When currents are in the same direction, the Force attracts
• When currents are in opposite directions, the Force repels
• The Force acting on a conductor is given as F = (µ₀ I₁ I₂ l) / (2πr)

Torque Acting on a Current-Carrying Conductor

• A rectangular loop PQRS with sides a and b carrying current I is placed in a uniform field B.
• The area vector A makes an angle θ with the field.
• Forces on arms QR and SP are equal, opposite, and collinear
• Forces on arms PQ and RS are equal and opposite but not collinear giving rise to torque.
• τ = NIAB sin θ where N is the number of turns.
• In vector form, τ = M × B, where M = NIA (magnetic dipole moment).
• The Torque = 0 if θ = 0°(plane normal to the field)
• The Torque is = NIAB if θ = 90°(plane parallel to the field)

Conversion of Galvanometer Into Ammeter and Voltmeter

• Galvanometer converted to ammeter by connecting a low shunt resistance R in parallel.
• Galvanometer converted to voltmeter by connecting a high resistance R in series.