   Chapter 19, Problem 29P

Chapter
Section
Textbook Problem

A wire with a mass of 1.00 g/cm is placed on a horizontal surface with a coefficient of friction of 0.200. The wire carries a current of 1.50 A eastward and moves horizontally to the north. What are the magnitude and the direction of the smallest vertical magnetic field that enables the wire to move in this fashion?

To determine
The magnitude of the magnetic field in the region, through which the current passes.

Explanation

Given info: The mass per unit length of the wire is 1.00gcm-1 . The wire is placed in a horizontal surface with a coefficient of friction 0.200 . The current in the wire is 1.50A directed eastward. The wire moves horizontally to the north.

Explanation:

The magnitude of the magnetic force on a current carrying wire is given by,

Fm=BILsinθ

• B is the magnetic field
• I is the current on the wire
• L is the length of the wire
• θ is the angle between the direction of current and the magnetic field

To find the minimum field to move the wire, the angle between the current and the magnetic field is 90° . Hence,

Fm=BIL

The kinetic frictional force on the wire is given by,

Ff=μsmg

• μs is the coefficient of friction
• m is the mass
• g is the free fall acceleration

To move the wire in this fashion the magnetic force has to be equal to that of the kinetic friction,

BIL=μsmg

On re-arrangement,

B=μsgI(mL)

Substitute 0.200 for μs , 1.00gcm-1 for mL , 1

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