Answered: A current I = 30 A is directed along…

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter22: Magnetic Forces And Magnetic Fields

Section: Chapter Questions

Problem 19P

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Question

A current I = 30 A is directed along the positive x-axis and perpendicular to a magnetic field. A magnetic force per unit length of 0.18 N/m acts on the conductor in the negative y-direction. Calculate the magnitude and direction of the magnetic field in the region through which the current passes.

magnitude

_____T

direction

A. +x direction

B. -x direction

C. +y direction

D. -y direction

E. +z direction

F. -z direction

Expert Solution

Step 1

The current $I = 30 A$ (along x axis)

Magnetic force per unit length $\frac{F}{L} = 0.18 N / m$ (in -y direction)

The magnetic force can be calculated be the formula given below

$F = I (\overset{⇀}{L} \times \overset{⇀}{B}) = ILB L is the length of the conductor B is the magnetic filed intensity$

$B = \frac{F}{LI} B = \frac{0.18 N / m}{30 A} B = 6 \times 10^{- 3} T$

Step 2

Using right hand rule to find the direction of cross product we get the magnetic field is in the -z direction.

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