What Is the Acceleration of the Metal Bar After Closing the Switch?

[kimberlyann9]

Homework Statement

A 2.60-N metal bar, 1.50 m long and having a resistance of 10.0 ohms, rests horizontally on conducting wires connecting it to the circuit shown in the figure . The bar is in a uniform, horizontal, 1.60-T magnetic field and is not attached to the wires in the circuit.

What is the acceleration of the bar just after the switch S is closed?
http://session.masteringphysics.com/problemAsset/1266176/3/27.74.jpg

Homework Equations

V=IR
Fbar=ILB
Fgravity=mg
Rseries=sum of R
Rparallel=(1/R)^-1

The Attempt at a Solution

I found the resistance of the parallel resistors(the bar and the 10ohm resistor) to be 5ohms
Then the Req=5ohms+25ohms=30ohms

Use that and the given voltage to find I.
I=4A

Use I to find Voltage of the bar.
V=20V

And then we can find I using the voltage(20V) and resistance of bar.

I=2A

F=Fbar-Fgravity
ma=ILB-mg

I get a=8.29 m*s^-2 but Mastering Physics says it's wrong. Where have I gone wrong?

 

[gneill]

Hi kimberlyann9, Welcome to Physics Forums.

Homework Statement

A 2.60-N metal bar, 1.50 m long and having a resistance of 10.0 ohms, rests horizontally on conducting wires connecting it to the circuit shown in the figure . The bar is in a uniform, horizontal, 1.60-T magnetic field and is not attached to the wires in the circuit.

What is the acceleration of the bar just after the switch S is closed?
http://session.masteringphysics.com/problemAsset/1266176/3/27.74.jpg

Homework Equations

V=IR
Fbar=ILB
Fgravity=mg
Rseries=sum of R
Rparallel=(1/R)^-1

The Attempt at a Solution

I found the resistance of the parallel resistors(the bar and the 10ohm resistor) to be 5ohms
Then the Req=5ohms+25ohms=30ohms

Use that and the given voltage to find I.
I=4A

Use I to find Voltage of the bar.
V=20V

And then we can find I using the voltage(20V) and resistance of bar.

I=2A

F=Fbar-Fgravity

Question: Why do you subtract the force due to gravity here? The bar is resting horizontally on the wire "rails", so it's supported. What is the direction of the force on the bar due to the magnetic field?

ma=ILB-mg

I get a=8.29 m*s^-2 but Mastering Physics says it's wrong. Where have I gone wrong?

 

[technician]

The diagram shows a VERTICAL magnetic field, not a horizontal field. This could be why you think you need to subtract mg.
Also did you notice the WEIGHT of the bar (2.60N) is given, you will need mass to use in the F =ma equation

 

[kimberlyann9]

Thanks guys I figured it out.

 

[asteriatic]

I would first check my calculations to make sure all of the steps were correct. If I am confident in my calculations, I would then consider other factors that could affect the acceleration of the bar, such as friction or air resistance. I would also double check the given values and equations to ensure they are accurate and appropriate for the problem. If all of these factors check out, I would then consider other physical phenomena that could be at play, such as electromagnetic induction or eddy currents, which could potentially affect the acceleration of the bar. I would also consider reaching out to colleagues or doing further research to see if there are any other factors I may have overlooked.