Answer:
Part a)
![F = 0](https://tex.z-dn.net/?f=F%20%3D%200)
Part b)
![F = 0.25 N](https://tex.z-dn.net/?f=F%20%3D%200.25%20N)
Part c)
![F = 0.25 N](https://tex.z-dn.net/?f=F%20%3D%200.25%20N)
Part d)
Net force on a closed loop in uniform magnetic field is always ZERO
![F_{net} = 0](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%200)
Explanation:
As we know that force on a current carrying wire is given as
![\vec F = i(\vec L \times \vec B)](https://tex.z-dn.net/?f=%5Cvec%20F%20%3D%20i%28%5Cvec%20L%20%5Ctimes%20%5Cvec%20B%29)
now we have
Part a)
current in side 166 cm and magnetic field is parallel
so we have
![F = i(\vec L \times \vec B)](https://tex.z-dn.net/?f=F%20%3D%20i%28%5Cvec%20L%20%5Ctimes%20%5Cvec%20B%29)
here we know that L and B is parallel to each other so
![F = 0](https://tex.z-dn.net/?f=F%20%3D%200)
Part b)
For 68.1 cm length wire we have
![F = iLB sin\theta](https://tex.z-dn.net/?f=F%20%3D%20iLB%20sin%5Ctheta)
here we know that
![cos\theta = \frac{68.1}{166}](https://tex.z-dn.net/?f=cos%5Ctheta%20%3D%20%5Cfrac%7B68.1%7D%7B166%7D)
![\theta = 65.8](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2065.8)
so we have
![F = (4.03)(0.681)(99.3 \times 10^{-3})sin65.8](https://tex.z-dn.net/?f=F%20%3D%20%284.03%29%280.681%29%2899.3%20%5Ctimes%2010%5E%7B-3%7D%29sin65.8)
![F = 0.25 N](https://tex.z-dn.net/?f=F%20%3D%200.25%20N)
Part c)
For 151 cm length wire we have
![F = iLB sin\phi](https://tex.z-dn.net/?f=F%20%3D%20iLB%20sin%5Cphi)
here we know that
![cos\phi = \frac{151}{166}](https://tex.z-dn.net/?f=cos%5Cphi%20%3D%20%5Cfrac%7B151%7D%7B166%7D)
![\theta = 24.5](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2024.5)
so we have
![F = (4.03)(1.51)(99.3 \times 10^{-3})sin24.5](https://tex.z-dn.net/?f=F%20%3D%20%284.03%29%281.51%29%2899.3%20%5Ctimes%2010%5E%7B-3%7D%29sin24.5)
![F = 0.25 N](https://tex.z-dn.net/?f=F%20%3D%200.25%20N)
Part d)
Net force on a closed loop in uniform magnetic field is always ZERO
![F_{net} = 0](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%200)
Answer:
![P_{avg} = 6.283*10^{-9} \ W](https://tex.z-dn.net/?f=P_%7Bavg%7D%20%3D%206.283%2A10%5E%7B-9%7D%20%5C%20W)
Explanation:
Given that;
I₀ = 9.55 A
f = 359 cycles/s
b = 72.2 cm
c = 32.5 cm
a = 80.2 cm
Using the formula;
![\phi = \frac{\mu_o Ic }{2 \pi} In (\frac{b+a}{b})](https://tex.z-dn.net/?f=%5Cphi%20%3D%20%5Cfrac%7B%5Cmu_o%20Ic%20%7D%7B2%20%5Cpi%7D%20In%20%28%5Cfrac%7Bb%2Ba%7D%7Bb%7D%29)
where;
![E= \frac{d \phi}{dt}](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7Bd%20%5Cphi%7D%7Bdt%7D)
![E = \frac{\mu_o}{2 \pi}c In (\frac{b+a}{a}) I_o \omega cos \omega t](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B%5Cmu_o%7D%7B2%20%5Cpi%7Dc%20In%20%28%5Cfrac%7Bb%2Ba%7D%7Ba%7D%29%20I_o%20%5Comega%20cos%20%5Comega%20t)
![E_{rms} = \frac { {\frac{\mu_o \ c}{2 \pi} In (\frac{b+a}{a}) I_o (2 \pi f)}}{\sqrt{2}}](https://tex.z-dn.net/?f=E_%7Brms%7D%20%3D%20%20%20%5Cfrac%20%7B%20%7B%5Cfrac%7B%5Cmu_o%20%5C%20c%7D%7B2%20%5Cpi%7D%20In%20%28%5Cfrac%7Bb%2Ba%7D%7Ba%7D%29%20I_o%20%282%20%5Cpi%20f%29%7D%7D%7B%5Csqrt%7B2%7D%7D)
Replacing our values into above equation; we have:
![E_{rms} = \frac { {\frac{4 \pi*10^{-7}*0.325}{2 \pi} In (\frac{72.2+80.2}{80.2}) *9.55 (2 \pi *359)}}{\sqrt{2}}](https://tex.z-dn.net/?f=E_%7Brms%7D%20%3D%20%20%20%5Cfrac%20%7B%20%7B%5Cfrac%7B4%20%5Cpi%2A10%5E%7B-7%7D%2A0.325%7D%7B2%20%5Cpi%7D%20In%20%28%5Cfrac%7B72.2%2B80.2%7D%7B80.2%7D%29%20%2A9.55%20%282%20%5Cpi%20%2A359%29%7D%7D%7B%5Csqrt%7B2%7D%7D)
![E_{rms} = \frac {8.98909588*10^{-4} }{\sqrt{2}}](https://tex.z-dn.net/?f=E_%7Brms%7D%20%3D%20%20%20%5Cfrac%20%7B8.98909588%2A10%5E%7B-4%7D%20%7D%7B%5Csqrt%7B2%7D%7D)
![E_{rms} = 6.356*10^{-4} \ V](https://tex.z-dn.net/?f=E_%7Brms%7D%20%3D%20%20%206.356%2A10%5E%7B-4%7D%20%5C%20V)
Then the
is calculated as:
![P_{avg} = \frac{E^2}{R}](https://tex.z-dn.net/?f=P_%7Bavg%7D%20%3D%20%5Cfrac%7BE%5E2%7D%7BR%7D)
![P_{avg} = \frac{(6.356*10^{-4})^2}{64.3}](https://tex.z-dn.net/?f=P_%7Bavg%7D%20%3D%20%5Cfrac%7B%286.356%2A10%5E%7B-4%7D%29%5E2%7D%7B64.3%7D)
![P_{avg} = 6.283*10^{-9} \ W](https://tex.z-dn.net/?f=P_%7Bavg%7D%20%3D%206.283%2A10%5E%7B-9%7D%20%5C%20W)
Answer:
9] V = D ÷ T
Take any distance value from the graph and its relevant time.
V = 4 ÷ 2
V = 2 m/s
[You will notice that any distance values with its time will give you 2 m/s as its speed. This means that speed is constant throughout.]
10] Take the distance value and its time for the highest peak of B.
V = 20 ÷ 2
V = 10 m/s