In a particular region there is a uniform current density of 15A/m2in the positive z direction. What is the value of B.dswhen that line integral is calculated along the three straight-line segments from (x, y, z) coordinates (4d, 0, 0) to (4d, 3d, 0) to (0, 0, 0) to (4d, 0, 0), where d=20cm?

Short Answer

Expert verified

The value of the Bdsis4.5×10-6T.m

Step by step solution

01

Listing the given quantities

  • The uniformcurrent density isj=15A/m2
  • Three straight-line segments fromx,y,z coordinates4d,0,0 to4d,3d,0 to 0,0,0to 4d,0,0where d=20cm.
02

Understanding the concept of Ampere’s circuital law

Ampere’s law states that,

B·ds=μ0i (i)

The line integral in this equation is evaluated around a closed-loop called an Amperian loop.The current ion the right side is thenetcurrent encircled by the loop.

We have to find the area enclosed by the closed-loopL. Using Ampere’s law, we can find the value ofB·ds.

03

Calculating the value of ∮B→·ds→

The area enclosed by the closed loopL is

A=124d3d=6d2=60.202=0.24m2

Thus, from equation (i),the value ofB.dsis,

B.ds=μ0i

B.ds=μ0jAand i=jA

B.ds=4π×10-715.00.24=4.5×10-6T·m

Therefore, the value of B.dsis 4.5×10-6T·m.

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