Multiple Choice Part A
- Using the curling right hand rule and Ampere’s Law we can determine that the magnetic field is pointing to the right, giving us answer C
- In this question we are looking for the distance after being given the current and magnetic field strength. We can solve this using the below formula, given in Lecture 13 - Sources of Magnetic Field.pdf. it is a derivation of the Biot-Savant Law
rearranging this, we get
Giving us answer B
- This question pertains the concepts of Induced EMF and Inductance. Looking at Physics 2 Lecture 15 and Faraday’s law, we can find the needed concepts.
- Electric fields can be generated by changing magnetic fields and vice versa
- Magnetic fields curl around current and vice versa.
Using the above two statements, we can deduce that there is a changing magnetic flux through A and C, but not B, therefore there is an induced EMF in A and C, but not B. Now, we need to find the direction of the induced EMF. We can use the right hand rule to find this after determining that the magnetic field is pointing out of the page for A, and into the page for B. The induced current would CCW for A, and CW for B. This yields us the answer D.
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A. is incorrect because that is a resistor, not an inductor B. is incorrect because that is a capacitor, not an inductor C. is incorrect because (fill reason here) D. Is correct because that is the definition of an inductor E. is incorrect because
Review DC circuits
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In this problem we are given the inductance and dimensions of a solenoid. The coil’s resistance is 9.9 ohms. The mutual inductance is 31 microhenry. At a given moment, the current in the solenoid 540 mA, and is decreasing at 2.5 A/s.
- These concepts were covered in Lecture 16 - Inductance and EMF.pdf and Ohm’s Law
- We can use the mutual inductance relation and Ohm’s Law to find the induced current. $$ \frac{V}{R} =I
3. Volume is known to be given by $LWH=V$
4. Calculating the magnetic field energy density we get:
```math
den= ((5E-5)^2/(2 * 1.26E-6)) J/m^3
vol= 3m * 4m * 2.4m
energy = vol * den
```
5. This corresponds to answer choice B.
7. AC RLC circuit analysis, this was covered in Lecture 18 - AC circuits and phasors.pdf and maybe Lecture 19 - Exponential analysis of ac circuits.pdf. The complex impedance of a RLC circuit is given by the following: $$ Z=\sqrt{R^{2} + \left( \omega L - \frac{1}{\omega C} \right)^{2}}
1. We can use the formula rearrange to find resistance as seen below: ```math omega = 2 pi 1000 l=25E-3 c=2E-6 z=200 r=sqrt(200^2 + (omega * l - (1/(omega * c))^2)) ``` 8. Series Circuit has a $50hz$ ac source, a $50\Omega$ resistor, a $0.5H$ inductor, and a $60\mu F$ capacitor. The RMS current is measured to be 3.1 A. 1. Formula for RMS $$ RMS_{I} = \frac{I}{\sqrt{ 2 }} \equiv RMS_{I} \ \cdot \ \sqrt{ 2 } = I2. Formula for inductance of a AC RLC Circuit:$$
Z=\sqrt{R^{2} + \left( \omega L - \frac{1}{\omega C} \right)^{2}}
3. Ohms Law: $$ V=IR4. Combining these,
1. $$
V=\sqrt{ 2RMS_{I}} \ \cdot \sqrt{R^{2} + \left( \omega L - \frac{1}{\omega C} \right)^{2}}
2. ```math f=50 omega = 2 * pi * f RMS = 3.1 R = 50 L = 0.5 C = 60E-6 V =RMS * sqrt(2) * sqrt(R^2 + (omega*L-(1/(omega*C)))^2) ``` 3. 506 volts is about 510, and therefore the answer is A 9. C, resistors don't lag or lead because they have no complex component. See [[Lecture 18 - AC circuits and phasors.pdf]] 10. C, resistors are not affected by frequency changes. Se [[Lecture 18 - AC circuits and phasors.pdf]] # Free Response Part B 1. This question is about a point particle moving through a magnetic field. [[Lecture 12 - Magnetic fields and forces.pdf]] 1. $\vec{F} = q\vec{v} \times \vec{B}$ 2. $\vec{F} = 2.00E-6 \ \cdot \ \langle 3000, 4000,0 \rangle \times \langle 1.5,0,0 \rangle$ 3. Cross product yields: 1. $\vec{F}=(2E-6) \cdot \langle 0,0,-6000 \rangle$ 2. $\vec{F} = \langle 0,0,-0.012 \rangle$ 4. The force vector is 0.012 newtons in the $-\hat{k}$ direction. 2. A current flows clockwise in a square loop that is halfway in a magnetic field of 0.25T pointing into the paper. 1. Looking at this problem, we can determine that the left and right side of the square will cancel each other out. Force on a straight wire in a uniform field is given by $\vec{F}_{B} = i\vec{L} \times \vec{B}$, and the magnitude of a force is given by $F=ILB$. We can ignore the lower wire, because its not in the magnetic field, so we just need to calculate the force on the upper wire. 2. $F=I\cdot L \cdot B$ 3. Force is therefore equal to ```math A = 0.2 B = 0.25 F = 0.3 I = F/(A * B ) ``` 3. Maximum voltage is given at natural frequency. Natural frequency is given by the following. 1. $\omega = \frac{1}{\sqrt{ LC }} \equiv f = \frac{1}{2\pi\sqrt{ LC }}$ 2. As seen below, the natural frequency is $1.591E7$ Hz. ```math C = 10E-6 L = 10E-4 f=1/(2 * pi * L * C) ``` 4. Circular wire loop radius 0.36 cm in x-z plane. Y-direction field changing at -0.0150 T/s. Find magnitude of induced electric field. 1. Using [[@SchoolFall 2023/School/Physics/Physics 1250/Concepts/Faraday's law]] we can determine the current induced in the loop.$$ \oint \vec{E} \cdot d\vec{l} = - \frac{d\Phi_{B}}{dt}2. Integrating over the loop, we get the equation $$
\mathcal{E} (2\pi r) = -\frac{d}{dt}(B \cdot A) \equiv \mathcal{E} = \frac{r}{2} \frac{dB}{dt}
3. Plugging in our known values for this, we get an induced electric field of 0.000027 V/m # Free response part C 1. The force is directed upwards because the particles charge is negative. 2. The magnitude of the force is given by $$ F = qV \times B1. Therefore, the magnitude of the force is 2.88e-17 N
3. The trajectory of the particle is going to travel along a curved path going towards the bottom. 4. We can use conservation of energy to solve this problem. $$ \frac{1}{2}mv^{2} = q\Delta V \implies \Delta V = \frac{1}{2}m \frac{v^{2}}{q}
1. This yields 0.11 volts 5. The magnetic field is given by $$ \vec{B} = \frac{\mu_{0}}{4\pi} \frac{q\vec{v} \times \vec{r}}{r^{3}}