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Chapter 29: Q59P (page 861)

What is the magnitude of the magnetic dipole moment μ→ of the solenoid described in Problem 51?

Short Answer

Expert verified

The magnitude of the magnetic dipole moment is0.46 A⋅m2.

Step by step solution

01

Listing the given quantities:

The length,l=25  cm=0.25m

The diameter,d=10  cm=0.10m

The number of turns,N=200

The current,i=0.29 A

02

Understanding the concept of magnetic field:

The magnitude of the magnetic dipole moment can be defined as the product of number of turns, current in that loop, and the area of that loop. Using this concept, you can calculate the magnitude of the magnetic dipole moment.

Formula:

The magnetic dipole moment is,

μ=NiA

Here, Nis the number of turns,i is the current,A is the area.

03

Calculations of the magnitude of the magnetic dipole moment:

To calculate the radius of the loop as

R=d2=0.10m2=0.05m

Now the magnetic dipole moment is given by,

μ=NiA=NiπR2=200×0.29A×3.14×(0.05m)2=0.46 A⋅m2

Hence, the magnitude of the magnetic dipole moment is 0.46 A⋅m2.

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Most popular questions from this chapter

In Figure, two long straight wires are perpendicular to the page and separated by distance d1=0.75cm. Wire 1 carries 6.5Ainto the page. What are (a) magnitude and (b) direction (into or out of the page) of the current in wire 2 if the net magnetic field due to the two currents is zero at point P located at distance d2=1.50cmfrom wire 2?

At a certain location in the Philippines, Earth’s magnetic field of 39μTis horizontal and directed due north. Suppose the net field is zero exactly8.0cmabove a long, straight, horizontal wire that carries a constant current. (a) What are the magnitude and (b) What is the direction of the current?

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Figure 29-67 shows a cross section across a diameter of a long cylindrical conductor of radius a=2.00cmcarrying uniform current 170A. What is the magnitude of the current’s magnetic field at radial distance (a) 0 (b) 1.00 cm, (c) 2.00 cm (wire’s surface), and (d) 4.00 cm?

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