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Chapter 29: Q. 33 (page 832)

What magnetic field strength and direction will levitate the 2.0gwire in FIGURE EX29.33?

Short Answer

Expert verified

The magnetic field in the wire is 0.13Tand it is out of the paper.

Step by step solution

01

Given information

We have given,

Current = 1.5A

Magnetic field region =10cm

We have to find the strength of magnetic field.

02

Simplify

We know the magnetic force is

F鈫�=i(L鈫�脳B鈫�)

To simplify it let us consider the magnetic field is out of the paper

F=iLBsin(90)0=iLBF=ILB

This force is acting upward and it is balancing the weight of the wire. then,

iLB=mg(1.5A)脳(0.1m)脳B=(0.002Kg)脳10B=0.13T

and the direction of magnetic field is out of the paper then only the force due to magnetic field will be able to balance the weight.

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Most popular questions from this chapter

What is the line integral of B鈫�between points i and f in FIGURE EX29.21?

What currents are needed to generate the magnetic field strengths of Table 29.1 at a point1.0cmfrom a long, straight wire?

What is the current direction in the wire of FIGURE Q29.4? Explain.

The magnetic field at the center of 1.0cmdiameter loop is 2.5mT.

a. What is the current in the loop?

b. A long straight wire carries the same current you found in part a. At what distance from the wire is the magnetic field2.5mT?

A conducting bar of length l and mass m rests at the left end of the two frictionless rails of length d in FIGURE P29.75. A uniform magnetic field of strength B points upward.

a. In which direction, into or out of the page, will a current through the conducting bar cause the bar to experience a force

to the right?

b. Find an expression for the bar鈥檚 speed as it leaves the rails at

the right end.
See all solutions

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