A solid cylindrical conducting shell of inner radius a 5.5 cm and outer radius b

Question

A solid cylindrical conducting shell of inner radius a 5.5 cm and outer radius b -7 cm has its axis aligned with the z axis as shown. It carries a uniformly distributed current l2- 5 A in the positive z-direction. An inifinte conducting wire is located along the z-axis and carries a current l-2.9 A in the negative z-direction , R(7074 707d) Pdo 1) What is By (P), the y-component of the magnetic field at point P, located a distance d 30 cm from the origin along the x-axis as shown? 1.77 Submit Your submissions:1.77x Computed value: 1.77 Feedback: Submitted: Thursday, March 29 at 6:13 PM 2) What is B dl where the integral is taken along the dotted path shown in the figure above: first from point P to point R at (x,y)- (0.707d, 0.707d), and then to point S at (x,y) - (0.6d, 0.6d) T-m Submit ) What is By(T), the y-component of the magnetic field at point T, located at (x,y) - (-5.9 cm,0), as shown? Submit

Explanation / Answer

from the given data

a = 5.5 cm

b = 7 cm

i1 = 2.9 A

i2 = 5 A

1. By(P) = ?

from ampere's law

B(2*pi*d) = mu*ienc

ienc = I2 - I1 = 2.1 A

hence

B = 2*10^-7*2.1/0.3 = 14*10^-7 T

2. in the given path

for the circular arc

B and dl are parallel, hence B.dl = b*dl*cos(0) = B*dl

for the radial part, B.dl = 0

hence

the integral is

B.dl = 14*10^-7*(pi*d/2) ( dl = theta*d = pi*d/2) = 6.597344*10^-7

3. at y = -5.9 cm, x = 0 cm

i.e. on a point on -ve y axis

magnetic field i along +x axis

hence By(T) = 0

4. from S to P

B.dl = - integration from the other path

because sum of these two paths should give 0 integral of bdl

hence

B.dl = -6.597*10^-7

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