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Questions: 1 2 3 4 5 6

Induction - Faraday's Law and Lenz' Law Solutions

  1. The diagram below shows a uniform magnetic field confined to a cylindrical region of space (seen end on). The magnitude of the field is B = 1.72 T. The circular cross-section of the region has a radius of 0.10 m. The magnetic field is also shown with three different square regions. The sides of the square regions are L1 = L2 = 0.30 m and L3 = 0.10 m.
    (a) Find the flux in each square region.
    (b) If the flux is increasing 0.025 T/s, what is the induced emf around the perimeter of each square region–

    (a) Magnetic flux is defined by Φm = BAcos(θ), where the angle θ is between the direction of the magnetic field B and a unit vector perpendicular to the area of interest. In this problem B is out of the page and so is the vector normal to the squares. Thus θ = 0 and cos(0) = 1.

    We need to be careful about area A. It is the area that contains magnetic field. For the first square, side L1, only the circle actually carries flux. So

    Φ1 = Bπr2 = (1.72 T)π(0.10 m)2 = 0.05404 T-m2 .

    For the second square only half the circle is enclosed, so

    Φ2 = ½Bπr2 = ½(0.0504 T-m2) = 0.02702 T-m2 .

    For the third small square, we need to ignore the field outside of the square. Hence here A is the size of the small square

    Φ3 = BL32  = (1.72 T)(0.10 m)2 = 0.0172 T-m2 .

    (b) Since only the magnetic field is changing, the emfs are

    ε1 =  –(dB/dt)πr2 = –(0.025 T/s)π(0.10 m)2 = –7.854 × 10-4 T-m2/s ,

    ε2 =  –(dB/dt)½πr2 = –(0.025 T/s)½π(0.10 m)2 = –3.9270 × 10-4 T-m2/s , and

    ε1 =  –(dB/dt) L32 = –(0.025 T/s) (0.10 m)2 = –2.50 × 10-4 T-m2/s .

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  2. 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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  3. 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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  4. 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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  5. 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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  6. 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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