- #1
Ozmahn
- 24
- 0
Homework Statement
A 2.60-N metal bar, 1.50 m long and having a resistance of 10.0 Ω, rests horizontally on conducting wires connecting it to the circuit shown in the figure(Figure 1) . The bar is in a uniform, horizontal, 1.60-T magnetic field and is not attached to the wires in the circuit.
Homework Equations
Fb=I LxB
Fg=mg
I=V/R
F=ma
The Attempt at a Solution
Resistance of bar and 10 ohm resistor are parallel, so combining them gives a ohm resistor. That is in series with the 25 ohm resistor, so the Req of the circuit is 30 ohms.
I=V/R=120/30=4 A
Using Kirchoff's laws, found that current passing through bar is 2 A.
Since current is moving clockwise, right hand rule says the force due to the magnetic field is moving the bar to the right.
The net force is Fb-Fg. Fg is given as 2.6N, Fb is
(2)(1.6)x(1.5)
Since L and B vectors are perpendicular, take the scalar product.
Fb=4.8N
To find m,
Fg=mg => Fg/g=m => 2.6/9.8=0.27 kg
To find acceleration, solve F=ma for a
a=F/m
So
a=(Fb-Fn)/m
a=(4.8-2.6)/.027
a=8.15
I got it wrong, and it says that the correct answer is 18.1 m/s^2. Any ideas as to where I went wrong? Thanks.