Chapter 30 IQ Questions - 30.1.1 Magnetic field at the...

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Unformatted text preview: 30.1.1 Magnetic field at the center of a current loop The loop has a current i and a radius r. Determine the direction of the magnetic field at the center. H' l. U th B. t S V3111 AB “0 i—z—ALXf ln : Se 9 10 — a. aw, = — . 43": 1' Extra: Based on the Biot—Savart law, show that the magnetic field at the center is given by B2u0if2r. Explanation: Taking the cross product, one finds that for any current segment along the circle, the corresponding AB at center always points into the paper. So B due to the entire current loop should also point into the paper. Extra: From the hint, the magnitude of the field at 0 due to a current segment irAG is AB:(p.D/41t) irABsinQOolrz. For the circle, one replaces AG by 211. This gives szuii2r. 30.1.2 Circular Arc A current loop consists of one circular arc subtended by an angle 0t, and two radial segments. There is a clockwise current i and the radius is r. Determine the direction of the magnetic field at the center, 0. H' U th B‘ S 1 AB H0 i Xf int: se e iot— avart aw, = — . 4:: 1-2 Extra: Based on the Biot—Savart law, show that the magnetic field at the center is given by B: uni 0c 1(4rtr). Explanation: Taking a cross product, one finds that for any current element along the arc, the corresponding AB at center always points into the paper. So B due to the entire arc should also point into the paper. Extra: From the hint, the magnitude of the field at 0 due to a current segment irAB is AB=(|.tU/4tt) irABsin900/r2. For the arc, one replaces AG by 0:. This gives B=poi 0t /(41tr). 30.1.4 Magnetic field due to a straight current segment Consider a. current segment CD with current i. Let B be the magnetic field vector at P due to this segment. Choose one: 1 2 l 3 l 4 dir. of B into into I out I out .3 Lei «fiuoi Lei «liuoi 4“ 4m. 4m 4m A A0 Hint: Consider iAy at y where sine = E ,—g] = —. Its l‘ r a M ' A0 contribution at P is AB = "—U—Zysine = fl —sin0. 4:: r 4a a Extra: Determine magnetic field vector at P due to CO. Explanation: By inspection iAyxr gives the direction of AB to be into the paper. So B at P due to CD is into the paper. Integrating from C to D, one obtains at P 3 4 - - B: T [Er—esineQ/EPU‘ m4 41': a 411a Extra: Since iAyxr gives the direction of AB to be into the paper, at P the magnetic field vector due to the segment CO should also be into the paper. For this case, the reader should verify that: B: ~/2p.Di/ (811a). 30.1.3 B due to a current loop C . b 1 a O A B Determine the direction of B at 0 due to the curent loop. 1 2 3 dirofB out‘ into ‘ BisO _ M0 in.fo Hint: = Extra: Find the magnitude of B. Explanation: B:BI+B2+Bs+B‘. At 0, 8228420. B1=(|J.o/4n )(iaJt/2)Ia2=|.toi/(8a), into the paper. B3:p.0if(8b), out of the paper. B1>B3. So resultant B is into the paper. Extra: B=(|.LUi/8)(1:’a-1/b). 3 30.3.1 B field due to 2 || wires 30.2.1 Magnetic force between "wire: ‘a O 3 i1 i2 ALP Two ||wires have currents i1 and i2. Dewmfine direcuon 0f the magnetic field BI! due to the Given 2 || wires where both currents enter into the paper. currenti at P- The distance of separation between them is 2a. Determine the direction of resultant field at P, where OP=2a. left Extra: Sketch a vector diagram, showing that the magnetic force vector, AFM, due to Bl on izAL is pointing to the left. right ‘ Etra:Sh thd'ti thP"d dtfth Explanation: By the RH-rule foralong wire, RHR #1, vine of 01)::V e "ea one a ls m epen en 0 e the direction of B1 at P is into the paper. See sketch. Extra: From the cross product rule, (RHR#2), the magnetic force AFM= iAL X B1 is pointing to the left. Explanation: Inspection on the sketch below shows that A APP’2m’2 — I3, and A APP”: I3, so A P’PP”:90°. B is along the negative x—axis. il Bl iAL 30.3.2 r—dependent current density _~‘-- P .__._'.«p .- a I Front view Top view Given: Current density szrz. Determine M of the Ampere’s law: M: [L B 'dS 2'10 iemir, where loop L is a circle with radius OP=a. 1 2 3 4 M ItaB/Z naB 2naB 41taB Extra: Based on iemi, =lJ21trdr, show that iemi, : nba4/2. Explanation: M = des= 2JtaB. Extra: lenc = 30.3.4 Field due to 4 ||—wires Consider 4 parallel wires shown in a 3d-view, and in a projected view from the right. The latter indicates that the wires are at four comers of a square, where each side has a length “a”. Sketch the field vectors due to the four wires at P, the center point, and the resultant vector B. | | 1 | 2 | 3 4 direction of B u down ri ht left Hint: For a longwire at a distance r, Bzuol/(2nr). Extra: Find the magnitude of B at P. Explanation: The situation at P is shown below. The vector sum gives the resultant field vector B, which is pointing in the upward direction. Extra: The sum of vertical projections of four vectors, 1 2 1 BB+BD, & BA+BC gives B = 4 cos45D "—0 = "D . “(7.5) 7“ 30.3.3 Flux through a square Consider the flux due to currenti in a long wire. Find the flux throu h the s uare area shown in the sketch. Hint: Att=BaAr, B=|.L°i/(2nr) Extra: Show I" = q.” b(a + b) equally divides the flux through the square. . “oiaznfl_ El a : = +bBad= xpanaonqfij'éz r2” b 1 b Extra: Let 311% = E £n(a Z Solving for r gives 30.3.5 Field due to current sheet A sheet of current is in the yz—plane at x:0. Linear current density L=dlzfdy=10 A/‘m. Determine the direction of B at P. 1 2 3 4 Dir of B J, _> 1~ (_ Extra: Consider the loop ABCDA, or L, where ' :la. lencir Show at P B=|J.°l/2. CB__ 1 a D E A"- Explanation: Right-hand-rule #1 together with the superposition principle implies that B is up. Extra: Ampere’s law says: the magneto-motive force over the complete Amperian loop ABCDA is given by: MLzBa+O+Ba+02uDla. Or BzuDJU2 30.4.1 Field outside of solenoid Strictly speaking B outside of an infinitely long solenoid is not zero. Find direction of B at oint P, OP=r. Hint: Sketch an Amperian loop through P. What is iemir? Extra: Show BP=u0i/(21tr). Explanation: Current from left to right is i. Applying the right-hand rule for a long wire, one finds that at P, B is along k. Extra: Applying Ampere’s law for the Amperian loop, which passes P, gives: 2an=tt0i, or BzuDi/(21tr). 30.5.1 B at one end of a long solenoid Given: A long solenoid which has a currenti and the linear number density (turns per length), n. Find: B A, the magetic field at the point A, located on the axis at the left end of the solenoid. l 1 l 2 l 3 l 4 l BA l pom | [Join | [Join/2 l [Join/2 l dir of B A Extra: Determine BC, magnetic field at C (at right end). Explanation: Assume the solenoid is long. Near the center BszR+BL=ZBR. By inspection, BAan, or BA: Bin/2:1L0in/2. (It is a special case of B=|,I,Din(sin11lz— sintpl)/2 given in the text. Here $500, and 1111 =—90°.) Extra: Again in the long solenoid approximation, by inspection, BC=BL. So BC is to the right, Bczpain/Z. 30.4.2 Ampere’s Law: Long Solenoid AN, AL=AB ...., Bout For a long solenoid with a current i, we assume the field inside B,,,, is uniform and constant and the field outside Bout, is 0. To evaluate the magnetic field inside: B=Bm, consider a rectangular Ampen'an loop of a height h and a Width AL. Let the number of Wires enclosed by the loop be AN. Determine the magnetomotive force (mmf) M along the loop ABCDA and the current enclosed: iencl. Hint: M involves an integral over the loop, Where the integrand at each segment is given by B-As=BAs cos 0. Extra: Show the field inside is B=B,n=p0iAN/AL. Explanation: The total magnetomotive force is given by: M=MAB+MBC+MCD+MDA. Since BJ_As, MBC=MDA=0, Since the magnetic field outside is 0, MCD=0. So M=MAB=+BAL. By inspection imlq'AN. Ans=3. Extra: With the Ampere’s law: M=pflimb BAL=ufliAN. or B=llgiANlAL. 30.8.1 Displacement Current Consider the set up shown in the sketch. Which of the following quantities isfare numerically equal to i? d dEA d¢E (WC 1. — , II. E —EE —, “LC— dtme 0 dt 0 dt (it [only I & II on] Extra: It is possible to have a situation where at some instant i¢0, and to have Q:Vc:¢E=0? Explanation: me=cvc. For a parallel plate system, EZQplate/(EUA), SO QplatE: Extra: Yes. We leave it as an exercise to the reader. 30.9.1 Magnetic dipole moment of an orbital electron V Consider an atom, where an electron is orbiting along a circular orbit with radius r, in a clockwise manner. The orbital speed is v. The electron has a. mass m, and a charge “— e”. Denote port, to be the corresponding magnetic dipole moment vector. Choose one: Hint: uorb =iAfi. Current is i=e/T, with period T=2nrfv. Extra: Verify that port = [2i]L,where L=mvr is the m orbital angular momentum. Explanation: By inspection, there is a counterclock current. The loop—magnet identity implies that port, is out of the paper. For a ciruclar orbit, the area is Aznrz, or new = iA= {%](nr2) = %. Extra: Since L=mvr, or VttL/m. Then [1.0m = {ZLJL m ...
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Chapter 30 IQ Questions - 30.1.1 Magnetic field at the...

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