EECE 380 Problems, Spring 2008

SOLUTION SOLUTION Determine the magnetic flux density at the center of a square loop of side L carrying a current I, as shown in the figure below. Considering the left side of the square, y we will apply Biot-Savart law to find the magnetic flux density due to a finite line. Then, we multiply the answer by 4 to get the magnetic flux density due to the square loop. Biot-Savart law: 4z\ R° R=ayr—ayy’ di'x R= ay dy’ X (ay — ayy’) =az,rdy’ =a MIL “anny (L/2y +7? In our case, 7 = L/2, and the magnetic flux density B due to the square loop is given by Mol * f rdy' B=|dB=a 7 ahalt } © an 2. (v2 +r? P? 2 2 =4xa, 2oIL =a, 22 pI AaLy(L/2) +(L/2) aL B total A square loop measuring 1.5 by 1.5 m carries a 7.5A steady current. The loop is put in the xz plane, the origin coinciding with a comer of the square, as shown in the figure. Calculate the B field at a point on the y axis 0.35m from the origin. We proceed by applying Biot-Savart law on every side of the square. (Same method as problem 1) Side Oc: Re=ay.r—a,.2’ di'* R=-a,dz’ x (ay.r—a,.2’) = ax (7 dz’) Mol + B,=a rdz x 3 4m 4 {2242 107 x7.5 1.5 B,. =a, —— = aa, x 2.087 0.35 V1.5? +0.35? rdz 1 v¥24er? “? IPar Side be: R=ax(-x)+ay.7—az.c di' x Bee as CGI eC r dx’) _fot i Ae ae * an yx? arr ae? Ml ¢ b “lor b =a, 2 2° > > 77a, 2 2° > > “Ar r+e VPsrPec? An rote be +ree? Note: 107 x7.5x1.5 1.5 0.35°41.5° 1.5? +4035? +15? 107 «7.5 0.35 1.5 0.354157 V1.5? 40.357 41.8 =(-a, «0.33 —a, x 0.077) uT y —a_ Note: | cdx _ ce x 2 2 > 2.2 A verse? rax _ sr x 2 2 2 272 2 2 2 xo+rt+e c+T xo+rt+e Proceeding in the same way for sides Oa and ab, we get B,, =a, x 2.087 nT = (-a, «0.077 —a, « 0.33) uT Therefore, the resultant B = 2.015a, —0.66a,+2.015a, uT PROBLEM 3 A semicircular loop of radius a carries a current J. The loop is situated in air. Find the magnetic field intensity at the center of the loop. SOLUTION Given that, for a circular loop carrying a current J, M,Ja? *2(h? +a?) where a is the radius of the circle and h is the distance from the plane of the loop to the point where B is measured. B=a 1 H=—B (inair) Ho 2 Ja“ In our case, we let h— 0, obtaining for a whole circle H = J/2a. For a semicircle, H will be half of this. or J/4a. Find a current distribution that produces a magnetic field of the form H = K sin x ay, where K is a constant. SOLUTION a, ay a, J=cuwlH =|— o —=Kcosxa, ox GO 0 Ksinx 0 An H due to a current source is given by H =(vcosax)a, + (v +e" hh, Describe the current density over the yz plane. SOLUTION J=cuwlH= a, =v +e j- ay <b +e )-a, S(ve0sas) =a, —e7a, —(cosax)a, For x = 0, this becomes J = a, — a, — a,, a constant vector of magnitude 3 A/mm’. PROBLEM 6 Determine the B-field corresponding to the magnetic vector potential A=ycosxay+ (vy + ea, SOLUTION . Since B=VxA, B=a, Hv + e*)-a, Llvee")-a, (ycosx) a oy -x . =a, +e a, —cosxa, PROBLEM 1 SOLUTION PROBLEM 2 SOLUTION PROBLEM 3 SOLUTION PROBLEM 4 SOLUTION A 0.5-m-long straight conductor carries 10 A of current in the positive z direction and is located 50 mm away from the z axis. The field in the region is given by B=-0.1sin(¢/2)a, (7). Find the force on the conductor. F =1f dlxB =Ifadl(a, xB)=JL,B,(a, xa, )=IL,B,a, F =10x0.5[-0.1sin(¢/2)p, =-0.5sin(¢/2)a, NV Relate the angles @ and @ in the figure below. There is no current sheet at the interface. From the boundary conditions for magnetostatic fields. we have Ay -Hy =F, =0 > Ay = Ay = Freose, =r cosé, (1) x Ay My and By = By > B, sin @, = B, sin é, (2) Divide (2) by (1): Li, tan@, = yw, tand, Refer to the figure in problem 2. Given B) = 1.2a, + O.8ay, 47={p and sy=15 4p. Determine Hj, and Bi. By continuity of H;, Hy = Hy = By2/tp = 1.2/15,zu0; hence 1.2 154 By continuity of B,, Baz = Bn2 = 0.8; hence B, = Bya, + 0.8a, The missing components are inferred from B; = 44H;: 1.2 0.8 B,=—— H,=-— 15 "Us l= a, +H,,a, A B-field given by B = 0.5ax + 0.Say T exists in free space. Calculate the energy density in the region. Bf =0.5? 0.8? =477 2 and . BE 1/2 5 ‘m = == MI!’ 2uy 2(4z x10") 87 v