Magnetism Forces caused by Magnetic Fields

Magnetic Fields Measured in Teslas (T) or Gauss (G) 10,000 G = 1 T Moving charge causes magnetic fields Remember the domain theory If moving charge causes magnetic fields, can a magnetic field affect a moving charge?

Force on a Moving Charge A moving charge experiences a force in a magnetic field F = qv x B The magnitude of the force is qvB sinθ The direction is determined by the Right Hand Rule

Apply RHR X v B Determine the direction of the force on the charge if the charge is a proton. F B Determine the direction of the velocity of the charge.

Apply RHR X vB F B F v

Force on a Current Carrying Wire A current carrying wire experiences a force in a magnetic field dF = I dL x dB F = IL x B The magnitude of the force is IlB sinθ The direction is determined by the RHR

Problem 1 A wire carrying a current of 2.0 A lies along the x-axis. The current flows in the positive x direction. A magnetic field of 1.2 T is parallel to the xy plane and makes an angle of 30º with the x-axis (pointing into the 1 st quadrant.) What is the force on a segment of wire 0.40 m long?

Solution F = IlB sinθ F = (2.0 A)(0.40 m)(1.2 T) sin 30º F = 0.48 N k

Problem 2 A wire carrying current of 1.5 A lies in a horizontal surface in the xy plane. One end of the wire is at the origin and the other is at (3m,4m). The wire follows an erratic path from one end to the other. A magnetic field of 0.15T directed vertically downward is present. What is the magnetic force acting on the wire?

Solution F = ILB sin θ F = 0.15T (1.5A) (3i-4j) F = √(3 2 – 4 2 ) F = 1.13 N This is the magnitude only!

Homework Problems Ch. 29 #1-5 and #33-38

F = qv x B Magnetic force on a charge at rest is zero. The magnetic force on a charge moving parallel or anti-parallel to the magnetic field is zero. Magnetic force, when non zero is perpendicular to velocity and magnetic field. Thus it contributes only to a radial acceleration and never performs work on an object or changes its speed (it can only change its direction).

Continued Magnetic force is proportional to the charge of a point charge and thus exerts forces only on charged objects. The direction can be obtained by the right- hand rule (positive charge only—reverse the direction of the velocity for a negative charge).

Tesla T = N/(Cm/s) T = N/Am

Change of direction F = qvB This force causes circular motion F = mv 2 /r mv 2 /r = qvB B = mv/qr Or m/q = Br/v (mass to charge ratio) Mass spectrometry allows us to find the mass to charge ratio

Three types of Motion When a charged particle moves into a magnetic field three types of motion can occur: Move parallel to B and experience no force Move perpendicular to B supplying a centripetal force Move at some angle to B and have components in the parallel and perpendicular direction producing helical or corkscrew pattern of motion

Force on a Current Carrying Wire F = qv x B F = e(Nv 0 ) x B N = # charges N = n(volume) = nlA Where l is the length of the wire and A is the cross-sectional area F = nlA ev o x B I = JA I = nev D A so F = Il x B dF = i dl x B F = ∫idl x B= i∫dl x B

Force on a Semicircle of Wire A wire bent into a semicircle of radius R forms a closed circuit and carries a current I. The wire lies in the xy plane and a uniform magnetic field is directed along the positive y axis. Find the magnitude and direction of the magnetic force acting on the straight portion of the wire and on the curved portion.

Answer F 1 acting on the straight portion has a magnitude of F 1 = IlBsinθ = I(2R)B sin 90° F 1 = 2IRB Direction out of the page or +z axis B

Continued F2 acting on the curved part—We must first write an equation for dF2 on the length ds dF2 = I ds X B = IB sin θ ds dF2 = IB sin θ ds s = R θ so ds = r d θ B

Continued θ = 0 to θ = π dF 2 = IRB sin θ d θ F 2 = IRB ∫ sin θ d θ From 0 to π F 2 = IRB (-cos θ) from 0 to π = -IRB (cos π – cos 0) = -IRB (-1-1) = 2IRB Direction: into the page or –z axis B

Torque on a Current Carrying Loop Torque = F x r Torque max = Frsinθ Torque max = F2 (b/2)sin 90 + F4(b/2)sin 90 = IaB(b/2) +IaB(b/2) = (IAB)/2 + (IAB)/2 = IAB Where A = ab I B b a

Continued Ends or part 1 and 3 Torque = F1 (a/2)sinθ + F3(a/2)sinθ Torque = IbB(a/2)sinθ + IbB(a/2)sinθ = IabBsinθ = IAB sinθ I 1 2 B b a 3 4

Magnetic moment Torque = I(A x B) μ = IA Torque = μ x B And U = - μ dot B A = Area Vector

Problem A rectangular coil of dimensions 5.40 cm x 8.50 cm consists of 25 turns of wire and carries a current of 15.0 mA. A 0.350T magnetic field is applied parallel to the plane of the loop. A) Calculate the magnitude of its magnetic dipole moment. B) What is the magnitude of the torque acting on the loop?

Answer μ = NIA = 25 (15 x A)( m)( m) = 1.72 x Am 2 Torque = 1.72 x Am 2 (0.350T) = 6.02x10 -4 Nm