Chapter 29: Q. 18 (page 831)

A small, square loop carries a $25 A$ current. The on-axis magnetic field strength $50 cm$ from the loop is $7.5 nT$ . What is the edge length of the square?

Short Answer

Expert verified

The edge length is $13.5 mm .$

Step by step solution

Given information

We have given,

Current = $25 A$

Distance = $50 cm$

Magnetic field = $7.5 nT$

We have to find the length of the edge of square.

Simplify

We know the magnetic field strength due to magnetic dipole moment

$B = \frac{μ_{0} \vec{μ}}{2 {πr}^{3}} \vec{μ} = \frac{2 {πr}^{3} B}{μ_{0}} \vec{μ} = \frac{2 π \times {(0.5)}^{3} \times 7.5 \times 10^{- 9}}{4 π \times 10^{- 7}} \vec{μ} = 4.6 mA . m^{2}$

Then magnetic moment for closed loop will be

$\vec{μ} = A I = L^{2} I L = \sqrt{\frac{\vec{μ}}{I}} L = \sqrt{\frac{4.6 \times 10^{- 3}}{25}} L = 13.5 m m$

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A small, square loop carries a $25 A$ current. The on-axis magnetic field strength $50 cm$ from the loop is $7.5 nT$ . What is the edge length of the square?

Short Answer

Step by step solution

Given information

Simplify

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A small, square loop carries a 25Acurrent. The on-axis magnetic field strength 50cm from the loop is 7.5nT. What is the edge length of the square?

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

Fluids

Astrophysics

Circular Motion and Gravitation

Magnetism and Electromagnetic Induction

Nuclear Physics

Oscillations

Study anywhere. Anytime. Across all devices.

A small, square loop carries a $25 A$ current. The on-axis magnetic field strength $50 cm$ from the loop is $7.5 nT$ . What is the edge length of the square?