The negative charges and positive charges in the wire are pushed in the opposite directions by the magnetic field
Answer:
Therefore energy is stored in the 1.0 mF capacitor is 5.56×10⁻⁹ J
Explanation:
Series capacitor: The ending point of a capacitor is the starting point of other capacitor.
If C₁ and C₂ are connected in series then the equivalent capacitance is C.
where ![\frac{1}{C} =\frac{1}{C_1}+\frac{1}{C_2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BC%7D%20%3D%5Cfrac%7B1%7D%7BC_1%7D%2B%5Cfrac%7B1%7D%7BC_2%7D)
Given that,
C₁ = 1.0 mF=1.0×10⁻³F and C₂ = 0.50mF=0.50×10⁻³F
If C is equivalent capacitance.
Then ![\frac{1}{C} =\frac{1}{1.0}+\frac{1}{0.5}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BC%7D%20%3D%5Cfrac%7B1%7D%7B1.0%7D%2B%5Cfrac%7B1%7D%7B0.5%7D)
![\Rightarrow \frac{1}{C} =\frac{3}{1}](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cfrac%7B1%7D%7BC%7D%20%3D%5Cfrac%7B3%7D%7B1%7D)
mF
Again given that the system is connected to a 100-v battery.
We know that
q=Cv
q= charge
C= capacitor
v= potential difference
Therefore
![q=(\frac{1}{3} \times 10^{-3}\times10) C](https://tex.z-dn.net/?f=q%3D%28%5Cfrac%7B1%7D%7B3%7D%20%5Ctimes%2010%5E%7B-3%7D%5Ctimes10%29%20C)
![=\frac{10^{-2}}{3} C](https://tex.z-dn.net/?f=%3D%5Cfrac%7B10%5E%7B-2%7D%7D%7B3%7D%20C)
The electrical potential energy stored in a capacitor can be expressed
![U=\frac{q^2C}{2}](https://tex.z-dn.net/?f=U%3D%5Cfrac%7Bq%5E2C%7D%7B2%7D)
q= charge
c=capacitance of a capacitor
Therefore energy is stored in the 1.0 mF capacitor is
![U=\frac{q^2C_1}{2}](https://tex.z-dn.net/?f=U%3D%5Cfrac%7Bq%5E2C_1%7D%7B2%7D)
![\Rightarrow U=\frac{(\frac{10^{-2}}{3})^2\times 10^{-3} }{2}](https://tex.z-dn.net/?f=%5CRightarrow%20U%3D%5Cfrac%7B%28%5Cfrac%7B10%5E%7B-2%7D%7D%7B3%7D%29%5E2%5Ctimes%2010%5E%7B-3%7D%20%7D%7B2%7D)
J
Explanation:
It is given that,
Distance between wires, d = 3.5 mm = 0.0035 m
Power of light bulb, P = 100 W
Potential difference, V = 120 V
(a) We need to find the force per unit length each wire of the cord exert on the other. It is given by :
![\dfrac{F}{l}=\dfrac{\mu_o I^2}{2\pi r}](https://tex.z-dn.net/?f=%5Cdfrac%7BF%7D%7Bl%7D%3D%5Cdfrac%7B%5Cmu_o%20I%5E2%7D%7B2%5Cpi%20r%7D)
Power, P = V × I
![I=\dfrac{P}{V}=\dfrac{100}{120}=0.83\ A](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7BP%7D%7BV%7D%3D%5Cdfrac%7B100%7D%7B120%7D%3D0.83%5C%20A)
This gives, ![\dfrac{F}{l}=\dfrac{4\pi \times 10^{-7}\times (0.83)^2}{2\pi \times 0.0035}](https://tex.z-dn.net/?f=%5Cdfrac%7BF%7D%7Bl%7D%3D%5Cdfrac%7B4%5Cpi%20%5Ctimes%2010%5E%7B-7%7D%5Ctimes%20%280.83%29%5E2%7D%7B2%5Cpi%20%5Ctimes%200.0035%7D)
![\dfrac{F}{l}=0.0000393\ N/m](https://tex.z-dn.net/?f=%5Cdfrac%7BF%7D%7Bl%7D%3D0.0000393%5C%20N%2Fm)
![\dfrac{F}{l}=3.93\times 10^{-5}\ N/m](https://tex.z-dn.net/?f=%5Cdfrac%7BF%7D%7Bl%7D%3D3.93%5Ctimes%2010%5E%7B-5%7D%5C%20N%2Fm)
(b) Since, the two wires carry equal currents in opposite directions. So, teh force is repulsive.
(c) This force is negligible.
Hence, this is the required solution.
B or D, one of these are the correct answer