   Chapter 19, Problem 26P

Chapter
Section
Textbook Problem

A wire having a mass per unit length of 0.500 g/cm carries a 2.00-A current horizontally to the south. What are the direction and magnitude of the minimum magnetic field needed to lift this wire vertically upward?

(a)

To determine
The magnitude of the minimum magnetic field required to lift the conductor.

Explanation

Given info: The magnitude of the current is 2.00A and is directed horizontally to the south. The mass per unit length of the wire is 0.500gcm-1 .

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

The magnetic force per unit length of the conductor is,

FmL=BIsinθ

The gravitational force per unit length will be,

FgL=(mL)g

• m is the mass of the wire
• g is the free fall acceleration

To lift the conductor vertically upward, the magnitude of the magnetic force per unit length should equal to the gravitational force per unit length and should be directed opposite to each other. Hence,

(mL)g=BIsinθ

Re-arrange for B ,

B=(mL)gIsinθ

Substitute 2.00A for I , 0

(b)

To determine
The direction of the minimum magnetic field required to lift the conductor.

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