Chapter 29: Q. 23 (page 831)

The value of the line integral of $\vec{B}$ around the closed path in FIGURE EX29.23 is $1.38 \times 10^{- 5} T . m$ . What are the direction (in or out of the page) and magnitude of $I_{3}$ ?

Short Answer

Expert verified

The current $I_{3} = 23 A$ and it is points into the paper.

Step by step solution

Given information

We have given,

line integral = $1.38 \times 10^{- 5} T . m$

We have to find the current $I_{3}$ and its flow of direction.

Simplify

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

$\oint \vec{B} . \vec{d S} = μ_{0} I$

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

then total current will be

$I = I_{2} + I_{3} I = - 12 A + I_{3} . . . . . . . . . . . . . . . . . . . . . . . (1)$

since other currents are out of closed curve.

since,

$\oint \vec{B} . \vec{d S} = μ_{0} I \oint \vec{B} . \vec{d S} = (4 π \times 10^{- 7} T . m / A) I I = \frac{\oint \vec{B} . \vec{d S}}{4 π \times 10^{- 7}} = \frac{1.38 \times 10^{- 5} T . m}{4 \times 3.14 \times 10^{- 7} T . m / A} I = 11 A$

using equation (1),

localid="1649421657359" $I = 11 A = - 12 A + I_{3} I_{3} = 23 A .$

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

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The value of the line integral of $\vec{B}$ around the closed path in FIGURE EX29.23 is $1.38 \times 10^{- 5} T . m$ . What are the direction (in or out of the page) and magnitude of $I_{3}$ ?

Short Answer

Step by step solution

Given information

Simplify

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The value of the line integral of B→around 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

Step by step solution

Given information

Simplify

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Fundamentals of Physics

Physics of Motion

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Further Mechanics and Thermal Physics

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Study anywhere. Anytime. Across all devices.

The value of the line integral of $\vec{B}$ around the closed path in FIGURE EX29.23 is $1.38 \times 10^{- 5} T . m$ . What are the direction (in or out of the page) and magnitude of $I_{3}$ ?