The two wires shown in the figure below are separated by d = 8.6 cm and carry cu

Question

The two wires shown in the figure below are separated by d = 8.6 cm and carry currents of I = 5.50 A in opposite directions.

(a) Find the magnitude and direction of the net magnetic field at a point midway between the wires.

(b) Find the magnitude and direction of the net magnetic field at point P1, 8.6 cm to the right of the wire on the right.

(c) Find the magnitude and direction of the net magnetic field at point P2, 2d = 17.2 cm to the left of the wire on the left.

magnitude _________________µT direction ---Select--- into the page, out of the page, toward the top of the page,toward the bottom of the page, toward the left side of the page, toward the right side of the page, no direction

Explanation / Answer

a) For point P to be midway(x=d/2) between the two wires

Magnetic field at P due to left wire,

B1 = u*I/(2*pi*(d/2)) <-----directed into the plane(direction is given by Right Hand Thumb rule)

where u = permittivity of space = 1.26*10^-6

I = 5.5 A

d = 8.6 cm = 0.086 m

Similarly, magnetic field at P due to right wire,

B2 = u*I/(2*pi*(d/2)) <----- directed into the plane

So, net magnetic field, Bent = B1 +B2 =  2*(u*I/(2*pi*(d/2))) = 2*u*I/(pi*d)

So, Bnet = 2*(1.26*10^-6)*5.5/(pi*0.086)

So, Bnet = 5.13*10^-5 T = 51.3 uT directed into the page<---------answer

b)

For P to be at P1,

Now, distance of left wire from P1, x1 = 8.6+8.6 cm = 2d = 17.2 cm = 0.172m

distance of left wiere from P2, x2 = 8.6 cm = 0.086 m

So, B1 = u*I/(2*pi*x1) = u*I/(2*pi*(2d))<----- directed into the plane

and B2 = u*I/(2*pi*x2) = u*I/(2*pi*d) <-----directed out of the page

So, Bnet = B2 - B1 <---- as both are directed in opposite directions

So, Bnet =  u*I/(2*pi*d) -  u*I/(2*pi*(2d)) =  u*I/(2*pi*(2d))

So, Bnet = u*I/(4*pi*d) = 1.26*10^-6*5.5/(4*pi*0.086)

= 6.41uT directed out of the page <---------answer

c)

Similarly,

B1 =  u*I/(2*pi*(2d)) <---- directed out of the page

B2 = u*I/(2*pi*3d) <----- directed into the plane

So, Bnet = B1 - B2 = u*I/(2*pi*(2d)) - u*I/(2*pi*3d)

So, Bnet = u*I/(4*pi*d) - u*i/(6*pi*d) = u*I/(12*pi*d)

So, Bnet = 1.26*10^-6*5.5/(12*pi*0.086)

= 2.14*10^-6 T = 2.14 uT directed out of the page <--------answer

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.