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Chapter 29: Q. 22 (page 831)

The value of the line integral of B鈫�around the closed path in FIGURE EX29.22 is 3.77脳10-5Tm. What isI3.?

Short Answer

Expert verified

The currentI3=7A.

Step by step solution

01

Given information

We have given,

line integral =3.77脳10-5Tm

We have to find the currentI3.

02

Simplify

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

鈭�B鈫�.dS鈫�=渭0I

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

then total current will be

I=I1+I2+I3I=2A-6A+I3I=-4A+I3.......................(1)

since

鈭�B鈫�.dS鈫�=渭0I鈭�B鈫�.dS鈫�=(4蟺脳10-7T.m/A)II=鈭�B鈫�.dS鈫�4蟺脳10-7=3.77脳10-5T.m4脳3.14脳10-7T.m/AI=3A

using equation (1),

I=-4A+I3=3AI3=7A.

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Most popular questions from this chapter

An electron moves in the magnetic fieldB鈬赌=0.50i^T with a speed of 1.0脳107m/sin the directions shown in FIGURE EX29.27. For each, what is magnetic force on the electron? Give your answers in component form.

a. Derive an expression for the magnetic field strength at distance d from the center of a straight wire of finite length l that carries current I.

b. Determine the field strength at the center of a current carrying square loop having sides of length 2R.

c. Compare your answer to part b to the field at the center of a circular loop of diameter 2R. Do so by computing the ratio BsquareBcircle.

What is the initial direction of deflection for the charged particles entering the magnetic fields shown in FIGURE

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?

A long, straight conducting wire of radius R has a nonuniform current density J=J0rR, where J0is a constant. The wire carries total current I.

a. Find an expression for J0in terms of I and R.

b. Find an expression for the magnetic field strength inside the wire at radius r.

c. At the boundary, r = R, does your solution match the known field outside a long, straight current-carrying wire?
See all solutions

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