Physics 1250 Homework 15

Question 1:

The figure below shows, several conductors that carry currents through the plane of the figure. $I_1=3.8A,I_2=5.6A,I_3=2.2A$

$$\oint \vec{B}\cdot dl={\mu }_0{I}_{enc}$$

Ampere’s Law Find the line integral for path A: $A=0$ because $I_{enc}=0$

Find the path integral for path b:

$$\oint \vec{B}\cdot dl={\mu }_0{I}_1=(1.25663706E-6)(3.8)=0.00000477522082770519$$

NB: It was negative b/c right hand rule, current is coming out of paper.

Find the path integral for path c:

$$\oint \vec{B}\cdot dl={\mu }_0(I_2-I_1)=(1.25663706E-6)(5.6-3.8)=0.00000226194670786035$$

Find the path integral for path d:

$$\oint \vec{B}\cdot dl={\mu }_0(I_3+I_2-I_1)=(1.25663706E-6)(2.2+5.6-3.8)=0.00000502654823968968$$

Question 2:

Toroidal solenoids This demonstrates amperes law quite well. For a solenoid of $N$ Turns and $r$ radius, the strength of a toroidal solenoid is given by $B=\frac{\mu NI}{2\pi r}$ . This is true because the current enclosed by the amperian surface surrounding the toroid is given by multiplying number of loops N by current I within each loop.

$$r=\frac{r_1+r_2}{2}$$ $$B_{16.1}=\frac{\mu NI}{2\pi r}=\frac{190\ast 8.4\mu }{2\pi \ast 0.161}$$

NB: I was a fucking nonce and didn’t realize that the phrase “from the center of the solenoid” meant from the center of the solenoid RADIALLY, not perpendicular to the solenoid.

Part A is zero because enclosed current is 0

Part C is also zero because the amperian surface encloses 0 net charge

Question 3:

solenoid

The Magnetic Field within a long solenoid is given below:

$$B=\frac{\mu NI}{L}$$

For our problem $N=600,L=0.15,I=8$ Therefore,

$$B=\frac{\mu \ast 600\ast 8}{0.15}=0.04021238591751741933$$

Question 4:

Hint 1: Find the current enclosed in an amperian loop Current enclosed is given by $IN$ because for N wires enclosed with I current in each. Hint 2: Use right hand rule to find that the direction of the magnetic field above the infinite sheet is in the $-\hat{i}$ direction Hint 3: Evaluate $\oint \vec{B}(\vec{r})\cdot \vec{d}l$ around the amperean loop $=2Ba$

Final Answer:

$$2Ba=I_{enc}{\mu }_0$$ $$-B=\frac{I_{enc}{\mu }_0}{2a}=\frac{NI{\mu }_0}{2a}\hat{i}$$