A rectangular loop of wire with sides H = 22 cm and W = 68 cm carries current I_

Question

A rectangular loop of wire with sides H = 22 cm and W = 68 cm carries current I_2 = 0.23 A. An infinite straight wire, located a distance L = 24 cm from segment ad of the loop as shown, carries current I_1 = 0.657 A in the positive y-direction. What is F_ad, x, the x-component of the force exerted by the infinite wire on segment ad of the loop? What is F_bc, x, the x-component of the force exerted by the infinite wire on segment bc of the loop?. What is F_net, y, the y-component of the net force exerted by the infinite wire on the loop? Another infinite straight wire, aligned with the y axis is now added at a distance 2L = 48 cm from segment bc of the loop as shown. A current, l_3, flows in this wire. The loop now experiences a net force of zero. What is the direction of l_3? along the positive y-direction along the negative y-direction What is the magnitude of l_3?

Explanation / Answer

magnetic field due to infinite wire B = uo*I1/(2*pi*r)

magnetic force F = I2*L*B

(1)

Fadx = I2*H*B

r = L

Fadx = + 0.23*0.22*4*pi*10^-7*0.657/(2*pi*0.24) = 2.77*10^-8 N

--------------------------

(2)

at r = L+W = 92 cm = 0.92 m

Fbcx = -I2*H*B

Fbcx = -0.23*0.22*4*pi*10^-7*0.657/(2*pi*0.92) = -7.227*10^-9 N

--------------------

(3)

Fnet y = 0

===================

along positive y direction

==============

Fadx due to I1 = Fadx due I3

I2*H*uo*I1/(2*pi*L) = I2*H*uo*I3/(2*pi*(2L+w))

I1/L = I3/(2L+w)

0.657/0.24 = I3/((2*0.24)+0.68)

I3 = 3.1755 A <<<<<----------answer

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---