Questions: 1 2 3 4 5 6

PHYS 2420 Induction Solutions

1. In the diagram below, a circular loop immersed in a magnetic field B. The field is oriented at an angle of θ = 25° to normal to the loop. The radius, r₀, of the loop is 10 cm. Take Ω = 2.5 rad/s and t = 3.5 s. Determine the emf and the direction of the current if

1. If B = 0.045 T;
2. If B = 0.045 T and r = 0.5r₀t²;
3. If B = −0.035t²+0.045;
4. If B = 0.050cos(Ωt);


Magnetic flux is defined by,

φ_m = òB · n dA = BAcos(θ ) .           (1)

From Faraday’s Law, the induced emf is

ε = −dφ_m/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) − 2πrB(dr/dt)cos(θ) + BAsin(θ)(dθ/dt) .            (4)

1. Since the magnetic field and all the other variables are constant, ε = 0.
2. Here r is changing, dr/dt = r₀t.

ε	= −2πrB(dr/dt)cos(θ)
	= −2πr02Bcos(25°)
	= −8.97 ´ 10-3 Volts

3. Here only B is changing, dB/dt = −0.07 T/s. We get

ε	= −(dB/dt)πr2cos(θ)
	= −(−0.07 T/s)π(0.10 m)2cos(25°)
	= +7.70 ´ 10-3 Volts

4. Again only B is changing,  dB/dt = −0.050Ωsin(Ωt). We get

ε	= −(dB/dt)πr2cos(θ)
	= (0.050Ωsin(Ωt))π(0.1 m)2cos(25°)
	= +2.22 ´ 10-3 Volts

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 the flux out through the loop is increasing, the current will be such to create flux into the page. This requires a clockwise current. If the flux out through the loop is decreasing, the counterclockwise current will be such to create flux out of the page.

In (b) the circle is increasing in size so the amount of flux up through the loop is increasing, so the current is CW. In (c), magnetic field is negative or into the page and getting more negative. This will induce a current which produces flux out of the page. Hence the current is CCW. In (d) it is harder to figure out what is going on. At t = 3.5 s the field is negative or into the page. At times just bigger than t = 3.5 s, B is even more negative. Hence we have downward increasing flux. Hence the induced current will be CCW. A simpler way to see this is that the sign of ε is the same as in (c) and (d) so the currents must be in the same direction.



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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 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.

1. Derive an expression for the magnetic flux φ_m as a function of x. Hints: The area of a triangle is one-half the base times the height. Consider similar triangles.
2. What is the magnitude of the induced emf at t = 0.30 s?
3. What is the direction of the induced emf at t = 0.30 s? Fully explain your reasoning.
4. If the resistance of the wire is 0.50 Ω, what is the current in the wire?


1. Magnetic flux is defined by,

φ_m = òB · n dA = 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/Ö3. Thus A = x²/Ö3.



Therefore

φ_m = Bx²/Ö3 .            (2)

2. The emf produced is given by Faraday’s Law

ε = -dφ_m/dt .            (3)

Differentiating (2) yields

ε = - (2/Ö3)Bx dx/dt .            (4)

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

ε = -(2/Ö3)Bv²t .            (5)

Evaluating εfor the given data yields

ε = -(2/Ö3)(0.355 T)(2.50 m/s)²(0.30 s) = -0.77 Volts.

3. 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.
4. Using Ohm’s Law, the current in the loop is

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



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5. Two parallel conducting rails are inclined at 30.0° 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 F_m = 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. Determine which of the following electric fields are non-conservative magnetically induced field. Determine the magnetic field which produced the non-conservative field.

1. E_x = x+y, E_y = −x+y, and E_z = −2z,
2. E_x = 2y, E_y = 2x+3z, and E_z = 3y,
3. E_x = x²−z², E_y = 2, and E_z = −2xz,
4. E_x =(y³−z³)cos(Ωt), E_y = (z³−x³)cos(Ωt), and E_z = (x³−y³)cos(Ωt).

Electric fields induced by magnetic fields obey Ñ·E_m = 0. The relationship between E_m and B is ¶B/¶t = Ñ´E_m. So we first take the cross product and then integrate.

(a)	Ñ·Em	= ¶Ex/¶x + ¶Ey/¶y + ¶Ez/¶z
		= ¶(x+y)/¶x + ¶(−x+y)/¶y + ¶(−2z)/¶z
		= 1 + 1 - 2
		= 0

So this electric field may have been magnetically induced.

Examining the cross product.

¶B/¶ t	= Ñ´Em

Integrating this result, the possible magnetic field has the form

B = C_xi + C_yj + (2t + C_z)k + C ,

Where C_x, C_y, C_z, and C are all constants.

(b)	Ñ·Em	= ¶Ex/¶x + ¶Ey/¶y + ¶Ez/¶z
		= ¶(2y)/¶x + ¶(−2x+3z)/¶y + ¶(3y)/¶z
		= 0 + 0 - 0
		= 0

So this electric field may have been magnetically induced.

Examining the cross product.

¶B/¶ t	= Ñ´Em

Integrating this result, the possible magnetic field has the form

B = C_xi + C_yj + C_zk + C ,

Where C_x, C_y, C_z, and C are all constants.

(c)	Ñ·Em	= ¶Ex/¶x + ¶Ey/¶y + ¶Ez/¶z
		= ¶(x2−z2)/¶x + ¶(2)/¶y + ¶(−2xz)/¶z
		= 2x + 0 - 2x
		= 0

So this electric field may have been magnetically induced.

Examining the cross product.

¶B/¶ t	= Ñ´Em

Integrating this result, the possible magnetic field has the form

B = C_xi + C_yj + (2t + C_z)k + C ,

Where C_x, C_y, C_z, and C are all constants.

(d)	Ñ·Em	= ¶Ex/¶x + ¶Ey/¶y + ¶Ez/¶z
		= [¶(y3- z3)/¶ x + ¶(z3− x3)/¶y + ¶ (x3 −y3)/¶ z]cos(Ωt)
		= [0 + 0 + 0] cos(Ωt)
		= 0

So this electric field may have been magnetically induced.

Examining the cross product.

¶B/¶ t	= Ñ´Em

Integrating this result, the possible magnetic field has the form

B = [3sin(Ωt)/Ω][i(y²+y²) + j(x²+z²) + k(x²+y²)] + C ,

Where C is a constants.

Questions?mike.coombes@kwantlen.ca