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Questions: 1 2 3 4 5 6

Induction - Faraday's Law and Lenz' Law Solutions

  1. In the diagram below, a circular loop is in a uniform magnetic field B = 0.045 T. The field is oriented at an angle of θ = 25° to normal to the loop. The radius of the loop is 10 cm.
    (a) Find the magnetic flux through the loop.
    (b) If the magnetic field decreases at a rate of 0.050 T/s, find the induced emf in the loop.
    (c) If, instead, the radius of the loop increases at 0.10 m/s, find the induced emf in the loop.
    (d) If direction of the magnetic field increases at the rate of 2.5 rad/s find the induced emf in the loop.
    (e) If all the above changes occur at the same time, find the induced emf in the loop.
    (f) Which way would a current flow in each case.

    (a) Magnetic flux is defined by,

    φm = BAcos(θ) . (1)

    Evaluating the flux for the given data,

    φm = BAcos(θ)
    = (0.45 T)π(0.1 m)2cos(25°)
    = 1.28 × 10-3 Tm2

    From Faraday's Law, the induced emf is

    ε = -dφm/dt . (2)

    Applying (2) to (1) yields

    ε = -(dB/dt)Acos(θ) - B(dA/dt)cos(θ) + BAsin(θ)(dθ/dt) . (3)

    Now the area of a circular loop is A = πr2, so equation (3) becomes

    ε = -(dB/dt)Acos(θ) - 2rB(dr/dt)cos(θ) + BAsin(θ)(dθ/dt). (4)

    (b) If only B is changing, dB/dt = -0.05 T/s and dr/dt = dθ/dt = 0. We get

    ε = -(dB/dt)Acos(θ)
    = -(-0.050 T/s)π(0.1 m)2cos(25°)
    = +1.42 × 10-3 Volts

    (c) If only r is changing, dr/dt = +0.10 m/s and dB/dt = dθ/dt = 0. We get

    ε = -2rB(dr/dt)cos(θ)
    = -2(0.1 m)(0.045 T)(0.010 m/s)cos(25°)
    = -2.56 × 10-3 Volts

    (d) If only is changing, dθ/dt = +2.5 rad/s and dB/dt = dr/dt = 0. We get

    ε = +BAsin(θ)(dθ/dt)
    = +(0.045 T)π(0.1 m)2sin(25°)(2.5 rad/s)
    = +1.49 × 10-3 Volts

    (e) If all variables are changing at once, the result is the sum of the answers from (b) to (d).

    ε = -(dB/dt)Acos(θ) - 2rB(dr/dt)cos(θ) + BAsin(θ)(dθ/dt)
    = +1.42 × 10-3 Volts - 2.56 × 10-3 Volts + 1.49 × 10-3 Volts
    = +3.5 × 10-4 Volts

    (f) We need to establish a coordinate system. In the diagram below, we show the coil as if we were looking down at it. The magnetic field is initially out of the page.

    Lenz's Law state the current will be such to oppose the change. If B out through the loop is increasing, the current will be such to create flux into the page. This requires a clockwise current. If B out through the loop is decreasing, the counterclockwise current will be such to create flux out of the page.

    In (b) B out of the page is decreasing, so the current is CCW. In (c), the area is increasing so the amount of flux though the loop is increasing. Hence the current is CW. In (d) the flux decreases as θ goes to 90°. Thus the current is CCW. In (e), the sign of is the same as in (b) and (d) so the current is in the same direction, CCW.

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  2. A single circular hoop moves with constant velocity through regions where uniform magnetic fields of the same magnitude are directed either into or out of the plane of the page as indicated below. Determined the direction of the induced current, if any, at each of the seven marked positions. HINT: sketch the flux as a function of position.

    At points (1), (3), (5), and (7) the flux is constant (zero) and no emf or current is produced.

    At (2) the flux out of the page through the loop is increasing. The emf and current are such to counter the growth by generating flux into the page. The current will be CW by the right hand rule.

    At (4) the flux out of the page through the loop is decreasing. The emf and current are such to counter the decrease by generating flux out of the page. The current will be CCW by the right hand rule.

    At (6) the flux into the page through the loop is decreasing. The emf and current are such to counter the decrease by generating flux into the page. The current will be CW by the right hand rule.

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  3. The DC-10 jet aircraft has a wingspan of 47 m. If such an aircraft is flying horizontally at 960 km/h at a place where the vertical component of the earth's magnetic field is 60 μT, what is the induced emf between its wingtips?

    We know that the emf produced in a conductor moving through a magnetic field at right angles is

    ε = lvB^
    = (47 m)(960 km/h × 1000 m/km × 1 h /3600 s) (60 × 10-6 T)
    = 0.75 Volts

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  4. In diagram (a) below, an equilateral triangle is just entering, at time t =0, a region of constant magnetic field B = 0.335 T into the page. In diagram (b) at some later time t > 0, the triangle has moved a distance x into the magnetic field. The triangle has sides of length L = 1.20 m long and is moving to the right at constant speed v = dx/dt = 2.50 m/s.
    (a) Derive an expression for the magnetic flux φm as a function of x. Hints: The area of a triangle is one-half the base times the height. Consider similar triangles.
    (b) What is the magnitude of the induced emf at t = 0.30 s?
    (c) What is the direction of the induced emf at t = 0.30 s? Fully explain your reasoning.
    (d) If the resistance of the wire is 0.50 Ω, what is the current in the wire?

    (a)Magnetic flux is defined by,

    φm = BAcos(θ) . (1)

    Here A is the area of the triangle that is in the magnetic field B. The triangle is perpendicular to the field so θ = 0 and cos(0) = 1. We will use a little geometry to find that area. Examining the diagram below, we see that A = ½bx. However we need to express b in terms of x. We have a right triangle so tan(30°) = b/2x and hence b = 2xtan(30°) = 2x/√3. Thus A = x2/√3.

    Therefore

    φm = Bx2/√3 . (3)

    (b) The emf produced is given by Faraday's Law

    ε = -dφm/dt . (3)

    Differentiating (2) yields

    ε = -(2/√3)Bx dx/dt . (4)

    Now v = dx/dt and x = vt for constant speed, so (4) becomes

    ε = -(2/√3)Bv2t . (5)

    Evaluating for the given data yields

    ε = -(2/√3)(0.355 T)(2.50 m/s)2(0.30 s) = -0.77 Volts .

    (b) The flux down through the loop is increasing. Lenz's Law says that the emf and current produced will counteract this by generating flux up through the loop. By the right hand rule, the emf and current are CCW.

    (c) Using Ohm's Law, the current in the loop is

    I = ε/R = 0.77 V / 0.50 = 1.54 Amps .

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  5. Two parallel conducting rails are inclined at 30 degrees to the horizontal, and are joined at the top by a length of copper wire; the rails and wire have negligible resistance. A 0.40 m long conducting rod of resistance 2.00 Ω slides without friction down the rails. Sliding through the magnetic field induces a current in the rod. The current-carrying rod then experiences a force from the external magnetic field. Assuming that the component of the magnetic field perpendicular to the incline, B^, points up, what magnitude must it have to ensure that the rod slides with a constant velocity of 5.00 m/s. The mass of the rod is 50.0 g. If the perpendicular component of the magnetic field pointed down what effect would this have? Why can we neglect the parallel component of the magnetic field, B||?

    Only B^ has an effect because B|| is in the same direction as the motion. Faraday's Law states that an emf is produced only when field lines are crossed.

    The magnitude of the produced emf is ε = vLB^, for a rod moving through a magnetic field. If we examine the diagram above, the area swept out by the rod is increasing. Therefore the flux through this area is also increasing out of the page. According to Lenz's Law, the emf produced must be CW to create a flux into the page. Since there is a conducting path there will be a CW current as well. Using Ohm's Law, the current will have a magnitude I = ε/R = vLB^/R .

    We also know that a current carrying wire moving through a magnetic field will experience a magnetic force Fm = ILB^ = v(LB^)2/R. Using the right hand rule, the force is directed up the incline.

    We are told that the velocity of the rod is constant so a = 0. Let's draw the free body diagram and apply Newton's Second Law.

    We get

    ΣFx = max ΣFy = may
    -Fm + mgcos(θ) = 0 N - mgsin(θ) = 0

    So we get

    v(LB^)2/R = mgcos(θ) .

    Solving for B^, we find

    B^ = [Rmgsin(θ) / vL2]½
    = [(2 Ω)(0.050 kg)(9.81 m/s2)sin(30°) / (5 m/s)(0.40 m)2]½
    = 0.78 T

    If the B were down instead of up, the current would circulate CCW. The magnetic force would still be up the incline, so the magnitude of B would remain the same.

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  6. The inductors in the circuit shown below are magnetically shielded from one another so that they do not produce flux in one another.
    (a) What is the equivalent inductance of the circuit?
    (b) When switch S is initially closed, what is the current produced by the battery?
    (c) When the switch has be been closed for a long time, what is the current produced by the battery?
    (d) When switch S is opened again, determine how long it takes the current through the resistor to drop to 85% of its maximum value.
    (e) At this point, what is the energy stored in each inductor?

    (a) The equivalent inductance is given by

    1/Leq = 1/L1 + 1/(L2+L3) = 1/30 + 1/50 = 1/18.75 .

    The equivalent inductance is Leq = 18.75 H.

    (b) With switch S closed initially, the inductors act as opens. The equivalent circuit is

    Using Ohm's Law,

    I0 = ε/(R1+R2) = 10 V / 100 Ω = 0.10 Amps .

    (c) After a long time, the inductors act as shorts

    Using Ohm's Law,

    I¥ = ε/R1 = 10 V / 75 Ω = 0.133 Amps .

    (d) The equivalent inductor is discharging through R2. According to our formula

    I(t) = I¥(1 - e-t/τ) , (1)

    where the time constant τ is

    τ = Leq/R2 = 18.75 H / 25 Ω = 0.75 s .

    We are looking for t such that I(t) = 0.85I¥ = 0.1133 A. Substituting into equation (1) yields

    0.85 = 1 - e-t/0.75 .

    Solving for t yields

    t = -(0.75 s)ln(0.15) = 1.42 s .

    (e) The energy stored in an inductor is given by U = ½LI2. Thus the equivalent inductor stores an energy of

    Eeq = ½Leq(0.85I¥)2 = ½(18.75 H)(0.1133 A)2 = 0.1204 J .

    For the inductors that make up the equivalent inductor we need to know how much current flows through each. The diagram below shows the current assignments.

    Since the arms are in parallel,

    L1dI1/dt = (L2 + L3)dI23/dt . (2)

    Using conservation of charge I = I1 + I23, this becomes

    L1dI1/dt = (L2 + L3)d(I - I1)/dt . (3)

    On rearrangement, this becomes

    dI1/dt = [(L2 + L3)/ (L1 + L2 + L3)]dI/dt . (4)

    Since we know the behaviour of I from (1), we integrate (4) and find

    I1 = [(L2 + L3) / (L1 + L2 + L3)]I = (50/80)(0.1133 A) = 0.0708 A .

    So the energy stored in inductor L1 is

    U1 = ½L1I12 = ½(30 H)(0.0708 A)2 = 0.0753 J .

    Similarly we find

    I23 = [L1 / (L1 + L2 + L3)]I = (30/80)(0.1133 A) = 0.0425 A .

    Thus the energy stored in L2 and L3 is

    U2 = U3 = ½L2I232 = ½(25 H)(0.0425 A)2 = 0.0226 J .

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