Physics EMF

Question

A metal rod of length 0.50 m is used to complete a circuit.  The circuit is perpendicular a magnetic field bit B = 0.15 T into the page.  If the resistance in the circuit is 3.0 Ohms and the rod is moved to the right at 2.0 m/s. what is the emf induce in the circuit, the current in the circuit, how large of force is needed to move the rod and what is the rate of work done by the motion of the bar as described?

Answer 1

E = induced emf

L = length of rod = 0.50 m

v = speed of rod = 2.0 m/s (assume positive to the right)

B = magnetic field pointing into page = -0.15 T (assume negative into page)

R = resistance = 3.0 Ω

 

 

E = -dΦ/dt = -d(BA)/dt = -BdA/dt 

 

A = L(vt) ⇒ dA/dt = Lv ⇒ 

 

E = -BLv = -(-0.15 T)(0.50 m)(2.0 m/s) = 0.15 volts

 

I = E/R = (0.15 volts / 3.0 Ω) amps = 0.05 amps

 

The direction of the current must be such that it produces a magnetic field that tries to cancel the change in flux as the rod moves to the right.  Given the constant magnetic field in this problem, the absolute value of the flux rises as the rod moves to the right.  However, the B lines are pointing into the page, so the flux is negative.  Therefore, the current must produce a magnetic field whose field lines point out of the page.  Then by the right hand rule, the current must flow in the counterclockwise direction to produce rising positive magnetic flux in order to counter rising negative magnetic flux.

 

F = magnitude of the force

 

F = |ILB| = (0.05 amps)(0.50 m)(0.15 T) = 3.75 x 10^-3 N; by the right hand rule, the magnetic force on the rod points toward the right

 

P = power = Fv = (3.75 x 10^-3 N)(2.0 m/s) = 7.5 x 10^-3 watts