A current is set up in a wire loop consisting of a semicircle of radius 49cm, a

Question

A current is set up in a wire loop consisting of a semicircle of radius 49cm, a smaller, concentric semicircle, and two radial straight lengths, all in the same plane. The figure below shows the arrangement but is not drawn to scale. The magnitude of the magnetic field produced at the center of curvature is 70µT. The smaller semicircle is then flipped over (rotated) until the loop is again entirely in the same plane. The magnetic field produced at the (same) center of curvature now has the magnitude of 32µT, and its direction is now reversed. What is the radius of the smaller semicircle in centimeters?

Hint: These two pictures will generate two equations, one for each picture. You'll end up with two unknowns, the current and the inner radius. Use algebra to solve for both of the unknowns.

Explanation / Answer

due to straight wire, field will be zero at centre.

and B1 due to larger and B2 is due to smaller semicircle.

for (A) :

B1 and B2 will be in same direction.

Bnet = B1 + B2 = (u0 I / 4R) + (u0 I / 4 r)

70 x 10^-6 = (4pi x 10^-7 x I / 4) (1/0.49 + 1/r)

222.82 = I ( 1/0.49 + 1/r) ......(i)

for (B):

B1 and B2 are in opposite direction.

Bnet = - B1 + B2 = - (u0 I / 4R) + (u0 I / 4 r)

32 x 10^-6 = (4pi x 10^-7 x I / 4) (- 1/0.49 + 1/r)

101.86 = I ( - 1/0.49 + 1/r) ......(ii)

(i)/(ii)

( 1/0.49 + 1/r) / ( - 1/0.49 + 1/r) = 2.1875

1/0.49 + 1/r = - 2.1875/0.49 + 2.1875/r

1.1875/r = 3.1875 / 0.49

r = 0.18 m Or 18 cm ...............Ans (inner radius)

101.86 = I ( -1/0.49 + 1/0.18)

I = 29.6 A .......ans

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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