The value of the line integral of Baround the closed path in FIGURE EX29.23 is 1.38×10-5T.m. What are the direction (in or out of the page) and magnitude of I3?

Short Answer

Expert verified

The current I3=23Aand it is points into the paper.

Step by step solution

01

Given information

We have given,

line integral =1.38×10-5T.m

We have to find the current I3and its flow of direction.

02

Simplify

Using ampere's law, for any closed path is given by,

B.dS=μ0I

Where I is a total enclosed current in the magnetic field curve.

then total current will be

I=I2+I3I=-12A+I3.......................(1)

since other currents are out of closed curve.

since,

B.dS=μ0IB.dS=(4π×10-7T.m/A)II=B.dS4π×10-7=1.38×10-5T.m4×3.14×10-7T.m/AI=11A

using equation (1),

localid="1649421657359" I=11A=-12A+I3I3=23A.

the positive sine indicates that current will be into the paper.

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