A solid cylindrical conducting shell of inner radius a = 5.8 cm and outer radius

Question

A solid cylindrical conducting shell of inner radius a = 5.8 cm and outer radius b = 7.4 cm has its axis aligned with the z-axis as shown. It carries a uniformly distributed current I_2 = 6.8 A in the positive z- direction. An infinite conducting wire is located along the z-axis and carries a current I_1 = 2 A in the negative z-direction. 1) What is B_y(P), the y-component of the magnetic field at Point P, located a distance d = 24 cm from the origin along the x-axis as shown? 2) What is integral^S_P 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). 3) What is B_y(T), the y-component of the magnetic field a point T, located at (x, y) = (-6.3 cm, 0), as shown? 4) What is integral^P_S B middot dl where the integral is taken on the straight line path from point S to point P as shown?

Explanation / Answer

1)

At P, field due to I2 and I1can be calculated as follows,

Let the field at be B.

Let us draw an Amperian loop at P of radius d. then we have

B.2d = µ0I = µ0(I1 + I2)

or, B = µ0(I1 + I2)/2d = (2X10-7T-m/A)(-2A + 6.8 A)/0.24m

or, B = 4X10-6 T

The direction of B at P is along +y-direction (using right hand thumb rule)

So, By(P) = 4X10-6 T

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2)

from P to R the field is By(P) = 4X10-6 T, and from R to S dl is perpendicular to the field, so B.ds will be zero from  R to S

from P to R, length of arc is L = 2d/8 = 0.1884 m

So, integral of B.dl is = (4X10-6 T)(0.1884 m)

or, integral of B.dl = 7.536X10-7 T-m

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3)

Since the current is uniformly distributed over the shell, then current enclosed in a cylinder of outer radius 6.3 cm and inner radius 5.8 cm is

I = 6.8A[(6.3cm)2 - (5.8cm)2]/[(7.4cm)2 - (5.8cm)2])

or, I = 1.95 A

Now, draw an Amperian loop at T of radius 6.3cm = 0.063m. then we have

B.2r = µ0I = µ0(I1 + I2)

or, B = µ0(I1 + I2)/2r = (2X10-7T-m/A)(2A - 1.95 A)/0.063m

or, B = 1.6X10-7T

So, By(T) = 1.6X10-7T

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4)

the integral value from S to P will be same if we take the from S to R and then R to P,

But from S to R integral is zero.

So integral from S to P is = -7.536X10-7 T-m (negative, as we found integral from P to R as 7.536X10-7 T-m).

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This concludes the answers. Check the answer and let me know if it's correct. If you need any more clarification or correction, feel free to ask.....

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