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Solutions – PHYS212 Final Exam - Practice Test
1 D
Let’s assume the direction of the current to be the positive direction. The induced EMF can be calculated using: €
EMF = − L
di dt
= −(1.5)(+ 3 ) = −4.5 V
Thus the induced EMF is 4.5 V and is in the opposite (negative) direction of the current. 2 C The current in the wire will induce a magnetic field going into the page on the right side of the wire. This causes an induced current in the loop such that the induced current creates a magnetic field opposing the one from the wire (meaning that induced magnetic field from the loop must travel up out of the page). This would happen if the current in the wire were moving counterclockwise according to the right-‐hand rule. 3 D The equation we’re looking for to connect these variables is: € ω =
LC
= 2 π f where f is the frequency mentioned in the problem. All we need do is solve for L: €
L =
4 π 2 f^2 C
= 5.07 × 10
− 7 H ≈ 0.5 μ H 4 B The equation for any circular loop is given by:
B = −
μ 0 i
2 R
= 2.0 × 10
− 5
T
However, since we are only dealing with half a loop, we will get half the amount of magnetic field, B/2, which equals 1.0 x 10-‐^5 T.
5 E
Using the equation for frequency: € ω =
LC
= 2 π f we notice that frequency is proportional to the inverse square root of the capacitance, C: € f ∝
C
which when we solve for capacitance we get: €
C ∝
f^2 In the problem we’re told that we double the frequency, and so according to the above proportion, doubling f causes C to become C/4. 6 C There are many ways to calculate power (which is the same thing as energy dissipation rate), but the one we need to use involves current and voltage: P = iV = (7.5 x 10-‐^3 )(120 x 10^3 ) = 900 Watts 7 B A bit of a trick question really. Since the circuit is in series, the current across all three devices will be the same, allowing the EMF to be in phase with the capacitor, so it’s frequency will be the same.
10 B
This one’s just straight from the packet. You should definitely be familiar with each of Maxwell’s Equations and what they specifically refer to. 11 E If there is no magnetic field inside the shell, then according to Gauss’ law there will no be no field inside the shell either, and thus the current will be zero. 12 A If we say that into the page is positive, we can evaluate the changes in magnetic flux to see what direction the induced magnetic field would need to point. A: No change in flux occurring so no induced current. B: ΔB = +1 – 0 = + This means the induced magnetic field must be -‐1 (out of the page) to oppose the existing change in flux. To make the induced magnetic field go out of the page, we curl our right hand fingers counter clockwise and our thumbs point up out of the page. C: Although there is a magnetic flux, it is constant, and thus no induced current. D: ΔB = 0 – +1 = -‐ 1 This means the induced magnetic field must be +1 (into the page) to oppose the existing change in flux. To make the induced magnetic field go into the page, we curl our right hand fingers clockwise and our thumbs point down into the page.
13 D
The equation for induced magnetic fields is:
B ⋅ ds = μ 0 ε 0
d Φ E
dt
∫ +^ μ 0 ienc
Notice that in this scenario the induced magnetic field will have the same sign as the changing electric flux. This means that the induced electric field which results from the induced magnetic field will point in the same direction as an increasing electric field (which would be away from the person). Since the electric field that causes this is not increasing but decreasing instead, we know that the induced magnetic field will do the opposite, to make sure that the induced electric field actually points towards the person. The only way an induced magnetic field could induce an electric field pointing towards the person is if it moves counterclockwise. Curl your right hand in a counterclockwise direction and notice that your thumb points towards you, which is the direction the induced electric field would have to follow. Note that in a question such as this, the magnetic field would always be perpendicular to the electric field, and so the answer could only be clockwise or counterclockwise. 14 E A somewhat interesting experiment, the magnet slows its decent due to the presence of an induced magnetic field that opposes the increasing flux of the magnet. This induced magnetic field is created by an induced current circling the cross-‐sectional circumference of the copper tube. 15 E Inside and outside the toroid, the magnetic field is zero, so only (a) and (e) are viable answers. We also should know that magnetic field will decrease as the radius increases within the shell of the toroid (but does not reach zero), and so (e) is the only option.
