   Chapter 19, Problem 70AP

Chapter
Section
Textbook Problem

A uniform horizontal wire with a linear mass density of 0.50 g/m carries a 2.0-A current. It is placed in a constant magnetic field with a strength of 4.0 × 10−3 T. The field is horizontal and perpendicular to the wire. As the wire moves upward starting from rest, (a) what is its acceleration and (b) how long does it take to rise 0.50 m? Neglect the magnetic field of Earth.

a)

To determine
The acceleration of the horizontal wire.

Explanation

Given info: The linear mass density of the uniform horizontal wire is 0.50gm-1 . Current in the wire is 2.0A . The strength of the magnetic field is 4.0×103T . The magnetic field is horizontal and perpendicular to the wire.

Explanation:

Consider upward as the positive direction.

The magnetic force acting on a wire is given by,

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

The gravitational force on the wire is working downward and is given by,

Fg=mg

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

The net vertical force on the wire will be given as,

F=FmFg=BILsinθmg

From Newton’s second law, the net force will be mass time acceleration,

ma=BILsinθmg

Re-arranging for the acceleration,

a=BILsinθmgm=BIsinθ(mL)g

Substitute 4

b)

To determine
The time it takes to rise 0.50m .

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