Force on a Current in a Magnetic Field

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SUMMARY

The discussion focuses on calculating the angle θ for a wire carrying a current of 42 A in a uniform magnetic field of 0.060 T, with a length of 0.20 m and mass of 0.065 kg. The key equation derived is arccos(mg / ILB) for θ, where m is the mass of the wire, g is the acceleration due to gravity, I is the current, L is the length of the wire, and B is the magnetic field strength. A common mistake identified was incorrectly canceling sin(θ) terms, leading to confusion in the solution process. The correct approach involves maintaining the distinction between the forces acting on the wire and their respective angles.

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  • Understanding of magnetic forces and the right-hand rule
  • Knowledge of equilibrium conditions in physics
  • Familiarity with trigonometric functions and their applications in physics
  • Basic understanding of Newton's second law
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  • Study the derivation of the Lorentz force equation F = ILB sin(θ)
  • Learn about the principles of static equilibrium in physics
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Help - Force on a Current in a Magnetic Field

Homework Statement


Magnetic and gravitational forces

A horizontal wire is hung from the ceiling of a room by two massless strings. A uniform magnetic field is directed from the ceiling to the floor. When a current exists in the wire, the wire swings upward and makes an angle with respect to the vertical (wire is in equilibrium).

p21-85.gif


The wire has a length of 0.20 m and a mass of 0.065 kg and carries a current of 42 A. The magnitude of the magnetic field is 0.060 T.

(e) What is the algebraic expression for the angle in the drawing? Express your answer in terms of the current I in the wire, the length L of the wire, the magnitude B of the magnetic field, the mass m of the wire, and the magnitude g of the acceleration due to gravity.

Homework Equations



F = ILB sin(theta)

The Attempt at a Solution



While this is a relatively easy problem, I still can't see what mistake I am making.

First, given that the wire is in equilibrium,
sigmacap.gif
Fx = 0 and
sigmacap.gif
Fy = 0

The three forces acting on the wire in the question are Tension in the two
Strings, the magnetic force from the magnetic field acting on the wire, and the force of gravity.

Using right hand rule, the magnetic force points to the right. It has to be perpendicular to the magnetic field and current (current replacing the velocity component in the right hand rule)

Knowing this, three equations are generated:

2T = Tension of two strings
the tension can be resolved into x and y components
x-component = 2Tsin(theta)
y-component = 2Tcos(theta)

Magnetic Force = ILBsin(theta) going to the right

Force of Gravity = mg going down

using Netwon's second law,
2Tcos(theta) - mg = 0
2T = mg / (cos(theta)) // solving for 2T using y-components

2Tsin(theta) - ILBsin(theta) = 0

2Tsin(theta) = ILBsin(theta) // solving for theta
2T = ILB
mg / (cos(theta)) = ILB // substitution from previous equation
mg / ILB = cos(theta)

arccos( mg / ILB ) = theta
This is the answer I arrived at for question (e), but according to my online problem set this answer is incorrect. If any insight could be given that would be great. Thanks!
 
Last edited by a moderator:
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There are two different angles that you have called theta. They are not the same.
 
apelling said:
There are two different angles that you have called theta. They are not the same.

Haha yeah, i just back from school today and figured it out during the day; one of my friends pointed it out. Silly me thinking you could cancel the sin(theta)'s. Thanks for the help.
 

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