Electromagnetic Induction: Analyzing a Sliding Wire in a Magnetic Field

Analyzing a Sliding Wire in a Magnetic Field

A wire of length L is sliding with velocity v_x > 0 without friction on rails that form part of a circuit with resistance R. A uniform magnetic field of magnitude B points into the paper.

1. [a] Find the induced emf E and the current I in the circuit, indicating the true positive direction of the current that flows in the sliding wire (up or down on the paper).

2. [b] Find the magnitude and direction of the magnetic force on the sliding wire in terms of B, L, v_x, and R.

3. [c] The wire has an initial velocity v_x(t = 0) = v₀, a mass m, and the only force acting on it is the force found in [b]. Show that the velocity obeys the equation:

   dv_x/ dt + v_x/ τ = 0, whose solution is v_x(t) = Ae^-t/τ

   and find τ in terms of m, R, L, and B. Also find the constant A.

4. [d] Since the velocity is a function of time, write the true current as a function of time, and find the heat dissipated in the resistor from t = 0 to t → ∞, for which you must do the integral:

   Heat = ∫₀^∞ i²(t)R dt

   and interpret how this is related to energy conservation.

⨂⨂⨂

⨂⨂⨂

⨂⨂ ⨂ v_x⨂⨂

L R⨂⨂⨂⨂⨂⨂⨂⨂⨂

B

Answers: Take v_x > 0 so the wire really is moving toward the right.

[a] For a Faraday loop, we use a curve that surrounds the rectangular area between the sliding wire and the resistor oriented in the counterclockwise direction. Then in the rectangular area, we use a normal n̂ out of the paper in the opposite direction as B⃗. This means that:

The magnetic flux is: Φ_B = B·n̂dA = –BdA = –B dA = –BLx.

The emf in the counterclockwise direction is:

E_{counterclockwise} = –dΦ_B/dt = BL(dx/dt) = BLv_x > 0, because here dx/dt = v_x.

⨂

⨂⨂

x

⨂

The counterclockwise current i_{counterclockwise} = E_{counterclockwise}/R is positive, so i_upward = BLv_x/R.

Physics 207 TEST 2 FORM 1 ANSWERS

June 15 2016

Dr. Huerta Phy 207 Test 2 FORM 1 ANSWERS

10:05 – 11:30 a. m., June 15 2016

[b] The magnetic force on the wire is:

F⃗_wire = i_upwardL⃗ × B⃗ = –i_upwardLBx̂ = -(LB)²v_x/R x̂,

that is, toward the left in the figure, opposite to the direction of motion of the wire.

[c] We use F⃗ = m⃗a for the wire. The force and the acceleration are toward the left, but we have defined ⃗v = v_xx̂ toward the right, so:

⃗a = dv⃗/dt = dv_x/dt x̂ = F_wire/m = -(LB)²v_x/Rm x̂ therefore dv_x/dt = -(B²L²/Rm)v_x.

For a solution, we try v(t) = v₀e^-t/τ, so:

dv/dt = -(B²L²/Rm)v gives –v₀/τ e^-t/τ = -(B²L²/Rm)v₀e^-t/τ, so τ = Rm/B²L².

[d] Given the solution to the differential equation, we see that the velocity is damped as:

v(t) = v₀e^-t/τ, where τ = Rm/B²L², and i(t) = BLv(t)/R = BLv₀e^-t/τ/R.

∫₀^∞ i²(t)R dt = ∫₀^∞ (B²L²/R²)v²(t)R dt = (B²L²/R)∫₀^∞ v₀²e^-2t/τdt = (B²L²/R)v₀²(-τ/2)e^-2t/τ|₀^∞ = (B²L²/R)v₀²(τ/2) = (B²L²/R)v₀²(Rm/2B²L²) = 1/2 mv₀².

or finally,

∫₀^∞ i²(t)R dt = 1/2 mv₀².

This means that the initial kinetic energy of the wire has been converted into an equal amount of heat in the resistor.