Dot Product Between Vectors - PDF Free Download

Dot Product etween Vectors Consider two vectors A = A x î + A y ĵ + A zˆk and = x î + y ĵ + zˆk. A = A cos φ = A A = A. A = A. A = A if A. A = 0 if A. A = (A x î + A y ĵ + A zˆk) (x î + y ĵ + zˆk) = A x x (î î) + A x y (î ĵ) + A x z (î ˆk) +A y x (ĵ î) + A y y (ĵ ĵ) + A y z (ĵ ˆk) +A z x (ˆk î) + A z y (ˆk ĵ) + A z z (ˆk ˆk). Use î î = ĵ ĵ = ˆk ˆk = 1, î ĵ = ĵ ˆk = ˆk î = 0. A = A x x + A y y + A z z. 7/10/2015 [tsl229 1/19]

Cross Product etween Vectors Consider two vectors A = A x î + A y ĵ + A zˆk and = x î + y ĵ + zˆk. A = A sin φ ˆn. A = A. A A = 0. A = A ˆn if A. A = 0 if A. A = (A x î + A y ĵ + A zˆk) (x î + y ĵ + zˆk) = A x x (î î) + A x y (î ĵ) + A x z (î ˆk) +A y x (ĵ î) + A y y (ĵ ĵ) + A y z (ĵ ˆk) +A z x (ˆk î) + A z y (ˆk ĵ) + A z z (ˆk ˆk). Use î î = ĵ ĵ = ˆk ˆk = 0, î ĵ = ˆk, ĵ ˆk = î, ˆk î = ĵ. A = (A y z A z y )î + (A z x A x z )ĵ + (A x y A y x )ˆk. 7/10/2015 [tsl230 2/19]

Magnetic Dipole Moment of Current Loop N: number of turns : current through wire A: area of loop ˆn: unit vector perpendicular to plane of loop µ = N Aˆn: magnetic dipole moment : magnetic field τ = µ : torque acting on current loop 7/10/2015 [tsl475 3/19]

Magnetic Moment of a Rotating Disk Consider a nonconducting disk of radius R with a uniform surface charge density σ. The disk rotates with angular velocity ω. Calculation of the magnetic moment µ: Total charge on disk: Q = σ(πr 2 ). Divide the disk into concentric rings of width dr. Period of rotation: T = 2π ω. Current within ring: d = dq T = σ(2πrdr) ω 2π = σωrdr. Magnetic moment of ring: dµ = d(πr 2 ) = πσωr 3 dr. Magnetic moment of disk: µ = Z R Vector relation: µ = π 4 σr4 ω = 1 4 QR2 ω. 0 πσωr 3 dr = π 4 σr4 ω. 7/10/2015 [tsl199 4/19]

Torque on Current Loop magnetic field: (horizontal) area of loop: A = ab unit vector to plane of loop: ˆn right-hand rule: ˆn points up. forces on sides a: F = a (vertical) forces on sides b: F = b (horizontal, not shown) torque: τ = Fbsin θ = A sin θ magnetic moment: µ = Aˆn torque (vector): τ = µ 7/10/2015 [tsl196 5/19]

Direct-Current Motor 7/10/2015 [tsl408 6/19]

Galvanometer Measuring direct currents. magnetic moment µ (along needle) magnetic field (toward right) torque τ = µ (into plane) 7/10/2015 [tsl409 7/19]

Electric Dipole in Uniform Electric Field Electric dipole moment: p = ql Torque exerted by electric field: τ = p E Potential energy: U = p E U(θ) = Z θ π/2 τ(θ)dθ = pe Z θ π/2 Note: τ(θ) and dθ have opposite sign. sin θdθ = pe cos θ q L p = ql +q τ E θ p 7/10/2015 [tsl197 8/19]

Magnetic Dipole in Uniform Magnetic Field Magnetic dipole moment: µ = Aˆn Torque exerted by magnetic field: τ = µ Potential energy: U = µ U(θ) = Z θ π/2 τ(θ)dθ = µ Z θ π/2 Note: τ(θ) and dθ have opposite sign. sin θdθ = µ cos θ n^ µ = An^. τ µ θ 7/10/2015 [tsl198 9/19]

Magnetic Force Application (7) The rectangular 20-turn loop of wire is 10cm high and 5cm wide. t carries a current = 0.1A and is hinged along one long side. t is mounted with its plane at an angle of 30 to the direction of a uniform magnetic field of magnitude = 0.50T. Calculate the magnetic moment µ of the loop. Calculate the torque τ acting on the loop about the hinge line. 30 o a = 10cm b = 5cm 7/10/2015 [tsl202 10/19]

Magnetic Force Application (4) A negatively charged basketball is thrown vertically up against the gravitational field g. Which direction of (a) a uniform electric field E, (b) a uniform magnetic field will give the ball a chance to find its way into the basket? (up/down/left/right/back/front) g V 7/10/2015 [tsl191 11/19]

Magnetic Force Application (6) An electric current flows through each of the letter-shaped wires in a region of uniform magnetic field pointing into the plane. Find the direction of the resultant magnetic force on each letter. NW W N NE E SW S SE 7/10/2015 [tsl193 12/19]

Magnetic Force Application (10) A triangular current loop is free to rotate around the vertical axis PQ. f a uniform magnetic field is switched on, will the corner R of the triangle start to move out of the plane, into the plane, or will it not move at all? Find the answer for a field pointing (a) up, (b) to the right, (c) into the plane. P R (a) (b) Q (c) 7/10/2015 [tsl205 13/19]

Hall Effect Method for dermining whether charge carriers are positively or negatively charged. Magnetic field pulls charge carriers to one side of conducting strip. Accumulation of charge carriers on that side and depletion on opposite side produce transverse electric field E. Transverse forces on charge carrier: F E = qe and F = qv d. n steady state forces are balanced: F E = F. Hall voltage in steady state: V H = Ew = v d w. positive charge carriers negative charge carriers 7/10/2015 [tsl201 14/19]

Charged Particle in Crossed Electric and Magnetic Fields (1) Release particle from rest. y Force: F = q( E + v ) E (1) F x = m dv x dt (2) F y = m dv y dt = qv y dv x dt = qv x + qe dv y dt = q m v y = q m v x + qe m Ansatz: v x (t) = w x cos(ω 0 t) + u x, v y (t) = w y sin(ω 0 t) + u y z m q x Substitute ansatz into (1) and (2) to find w x, w y, u x, u y, ω 0. (1) ω 0 w x sin(ω 0 t) = q m w y sin(ω 0 t) q m u y (2) ω 0 w y cos(ω 0 t) = q m w x cos(ω 0 t) + q m u x + qe m u y = 0, u x = E, ω 0 = q m, w x = w y w nitial condition: v x (0) = v y (0) = 0 w = E 7/10/2015 [tsl208 15/19]

Charged Particle in Crossed Electric and Magnetic Fields (2) Solution for velocity of particle: v x (t) = E» «qt cos 1, v y (t) = E «qt m sin m Solution for position of particle: x(t) = E Z t» «qt cos 1 dt = Em «qt 0 m q 2 sin m y(t) = E Z t «qt sin dt = Em» «qt m q 2 1 cos m 0 Et Path of particle in (x, y)-plane: cycloid t = 2πm q y 2mE q 2 m E q x 2πmE q 2 7/10/2015 [tsl209 16/19]

ntermediate Exam : Problem #1 (Spring 07) Consider a rectangular conducting loop in the xy-plane with a counterclockwise current = 7A in a uniform magnetic field = 3Tî. (a) Find the magnetic moment µ (magnitude and direction) of the loop. (b) Find the force F (magnitude and direction) acting on the side ab of the rectangle. (c) Find the torque τ (magnitude and direction) acting on the loop. y 9m b 5m a x z 7/10/2015 [tsl365 17/19]

Unit Exam : Problem #1 (Spring 08) Consider two circular currents 1 = 3A at radius r 1 = 2m and 2 = 5A at radius r 2 = 4m in the directions shown. (a) Find magnitude and direction (, ) of the resultant magnetic field at the center. (b) Find magnitude µ and direction (, ) of the magnetic dipole moment generated by the two currents. 2 1 r 2 r 1 7/10/2015 [tsl381 18/19]

Unit Exam : Problem #1 (Spring 09) A triangular conducting loop in the yz-plane with a counterclockwise current = 3A is free to rotate about the axis PQ. A uniform magnetic field = 0.5Tˆk is present. (a) Find the magnetic moment µ (magnitude and direction) of the triangle. (b) Find the magnetic torque τ (magnitude and direction) acting on the triangle. (c) Find the magnetic force F H (magnitude and direction) acting on the long side (hypotenuse) of the triangle. (d) Find the force F R (magnitude and direction) that must be applied to the corner R to keep the triangle from rotating. z 8m P Q 8m x R y 7/10/2015 [tsl395 19/19]