How Does a Conducting Bar on Rails Behave in a Magnetic Field?

[krh68]

A conducting bar of length L = 21.2 cm and mass M = 60.0 g lies across a pair of conducting rails. The contact friction between the bar and the rails is negligible, but there is a resistor at one end with a value R = 30.0 Ohms. Initially the rod is given an initial speed of v0 = 64.0 meters per second. There is a uniform magnetic field perpendicular to the plane containing the rod and rails of magnitude B = 1.3 T.
What is the speed of the rod at time t = 26.068 s?

I know:
v=v0 + at
F=ma
F=iLB
i=(emf)/R
emf = dflux/dt
flux = BA

I know I need to solve for the area to get the flux and the length (L) is constant while the width is changing but I don't understand how to get the integral or set up the integral for the width. Please help ASAP.

 

[kuruman]

v = v₀+at is out of the picture. This is not a constant acceleration situation. All your other equations are relevant.
You need to set up a differential equation and solve it. Start with

F = m (dv/dt)

Replace F with iLB and then replace i with (1/R)(dΦ/dt). The expression for dΦ/dt is proportional to v. So you end up with the differential equation that is essentially

dv/dt = (const)v

You should be able to find what "const" is and to integrate the above equation.

 

[krh68]

Okay, so dv/dt = v(const) or dv/dt = v(LB/mR)
I still don't know how to solve for v.
When I integrate dv/dt, do I get r(LB/mR)? and if so, what is r?

 

[kuruman]

Okay, so dv/dt = v(const) or dv/dt = v(LB/mR)

This is incorrect. Please show how you got it, then I can point out where you went wrong.

I still don't know how to solve for v.

Worry about that later. First get the correct expression for dv/dt.

When I integrate dv/dt, do I get r(LB/mR)?

No, you do not.

and if so, what is r?

I don't know, but r it appears in your expression above. You made it up so you should know what it represents.

 

[rwisz]

Based on the given information, the speed of the rod at time t = 26.068 s can be determined by using the equations for motion and electromotive force (emf). First, we can use the equation v=v0 + at to calculate the acceleration of the rod. Since the initial speed (v0) and time (t) are given, we can solve for the acceleration (a).

v=v0 + at
a=(v-v0)/t
a=(0-64.0 m/s)/(26.068 s)
a=-2.4575 m/s^2

Next, we can use the equation F=ma to calculate the force acting on the rod. Since the mass (M) is given, we can solve for the force (F).

F=ma
F=(60.0 g)(-2.4575 m/s^2)
F=-0.14745 N

Then, we can use the equation F=iLB to calculate the current (i) in the circuit. Since the force (F) and length (L) are known, we can solve for the current.

F=iLB
i=F/(LB)
i=(-0.14745 N)/(1.3 T)(0.212 m)
i=-0.546 A

Next, we can use the equation i=(emf)/R to calculate the emf. Since the current (i) and resistance (R) are known, we can solve for the emf.

i=(emf)/R
emf=iR
emf=(-0.546 A)(30.0 Ohms)
emf=-16.38 V

Finally, we can use the equation emf = dflux/dt to calculate the change in flux over time. Since the emf (16.38 V) and time (26.068 s) are known, we can solve for the change in flux (dflux).

emf = dflux/dt
dflux = emf*dt
dflux = (16.38 V)(26.068 s)
dflux = 426.24624 V*s

Since the flux (flux) is equal to the product of the magnetic field (B) and the area (A), we can solve for the area (A).

flux = BA
A = flux/B
A = (426.24624 V*s)/(