Question

You know that the net magnetic force on the (closed) current loop in the uniform magnetic field is zero. As shown in the figure below, the rectangular thin wire loop running current I₂=8.97 A is, however, in the non-uniform magnetic field produced by a long straight wire carrying current I₁=13.4 A. The rectangular loop and the source wire are in the same (xy)-plane with the geometrical dimensions illustrated in the figure and equal to a=0.68 cm, b=1.7 cm and c=53.5 cm.

Find Cartesian components F_x and F_y of the net magnetic force F exerted on the rectangular loop:

F_x= _____ mN,

F_y= _____ mN.

What is the magnitude of the flux
Φ_B =

	BndA

of the magnetic field of the source wire through the surface bounded by the rectangular loop?:

|Φ_B|= _____ G·cm².

(Since the magnetic field is not uniform over the surface of the loop, you do need to integrate over the area.)

You know that the net magnetic force on the (closed) current loop in the uniform magnetic field is zero. As shown in the figure below, the rectangular thin wire loop running current 12=8.97 A is, however, in the non-uniform magnetic field produced by a long straight wire carrying current 11-13.4 A. The rectangular loop and the source wire are in the same (xy)-plane with the geometrical dimensions illustrated in the figure and equal to a=0.68 cm, b=1.7 cm and c=53.5 cm. 12 с a b Find Cartesian components Fx and Fy of the net magnetic force F exerted on the rectangular loop: mN, Fy= mN. What is the magnitude of the flux BndA of the magnetic field of the source wire through the surface bounded by the rectangular loop?: = G.cm2 (Since the magnetic field is not uniform over the surface of the loop, you do need to integrate over the area.)

Answer 1

The solution is given below.

the side pQ and sk are perpendicular to the Cansining wine I,,.., force on them is zeror Correur Iz Force on side Ps is fps 2 I, I2 Xc no Чt alorgan 71 dre IR 4TX10 2x 13.4 x 8.97 X0.55 0.0068 -a k b & C= 53.5cm = 0.55m a=0.68 en -0.0068 m 1.7 cm = 0.017 m 1.891 3 Xo fps +1.891 mN Forree on side Q R is 2 I, I2 Calore -x) Un ath 40x107 2% 13.4*8.92 0.535 0.006 8+ 0.017 3 = 5.403 410 N FOR = 0.5403 410 N -0.5403 MN Net forreo exerted on Eps - FOR = (1.891 0.5403 ) fx = 1.350 mN There is component the wop along & axis iffa omn Answer: x = 1.350 mN MN. magnetic field at a distance r from I, is B = us 2 I YM magnetic flux for sisnall do portion of rectangular wap d = Biedro area dA=cdr 2 Il e da عنا 211 total that the police st mo ati clar] 2I, C luca the 21 c um 0 B 4

x 2 x 13.4 x 0.535 PBS 2I, M. -7 4 ITX10 enfu a 0.68 1.31378x106 Tm2 = 1.31378x101x10 G.com an2 Now IT=109 Ime = 104 en - 131.378 Ganz Answer: 146 1= 131-378 G. con

Thank you.