Answer:
The induced emf in the loop is ![7.35\times 10^{-4}\ V](https://tex.z-dn.net/?f=7.35%5Ctimes%2010%5E%7B-4%7D%5C%20V)
Explanation:
Given that,
Length of the wire, L = 1.22 m
It changes its shape is changed from square to circular. Then the side of square be its circumference, 4a = L
4a = 1.22
a = 0.305 m
Area of square, ![A=a^2=(0.305)^2=0.0930\ m^2](https://tex.z-dn.net/?f=A%3Da%5E2%3D%280.305%29%5E2%3D0.0930%5C%20m%5E2)
Circumference of the loop,
![C=2\pi r=L\\\\r=\dfrac{L}{2\pi}\\\\r=\dfrac{1.22}{2\pi}=0.194\ m](https://tex.z-dn.net/?f=C%3D2%5Cpi%20r%3DL%5C%5C%5C%5Cr%3D%5Cdfrac%7BL%7D%7B2%5Cpi%7D%5C%5C%5C%5Cr%3D%5Cdfrac%7B1.22%7D%7B2%5Cpi%7D%3D0.194%5C%20m)
Area of circle,
![A'=\pi r^2\\A'=\pi (0.194)^2\\\\A'=0.118\ m^2](https://tex.z-dn.net/?f=A%27%3D%5Cpi%20r%5E2%5C%5CA%27%3D%5Cpi%20%280.194%29%5E2%5C%5C%5C%5CA%27%3D0.118%5C%20m%5E2)
The induced emf is given by :
![\epsilon=\dfrac{\d\phi}{dt}\\\\\epsilon=\dfrac{\d(BA)}{dt}\\\\\epsilon=B\dfrac{A'-A}{t}\\\\\epsilon=0.125 \times \dfrac{0.118-0.0930}{4.25}\\\\\epsilon=7.35\times 10^{-4}\ V](https://tex.z-dn.net/?f=%5Cepsilon%3D%5Cdfrac%7B%5Cd%5Cphi%7D%7Bdt%7D%5C%5C%5C%5C%5Cepsilon%3D%5Cdfrac%7B%5Cd%28BA%29%7D%7Bdt%7D%5C%5C%5C%5C%5Cepsilon%3DB%5Cdfrac%7BA%27-A%7D%7Bt%7D%5C%5C%5C%5C%5Cepsilon%3D0.125%20%5Ctimes%20%5Cdfrac%7B0.118-0.0930%7D%7B4.25%7D%5C%5C%5C%5C%5Cepsilon%3D7.35%5Ctimes%2010%5E%7B-4%7D%5C%20V)
So, the induced emf in the loop is ![7.35\times 10^{-4}\ V](https://tex.z-dn.net/?f=7.35%5Ctimes%2010%5E%7B-4%7D%5C%20V)
Elastic potential energy is equal to the force times the distance of movement. Elastic potential energy = force x distance of displacement. Because the force is = spring constant x displacement, then the Elastic potential energy = spring constant x displacement squared.
Answer:
0
Explanation:
F1 = G•2.2•4.66/3² (pointed right)
F2 = G•2.2•4.66/3² (pointed left)
subtract the two to get zero
Answer:
Explanation:
Given
Charge of first Particle ![q_1=+12\ \mu C](https://tex.z-dn.net/?f=q_1%3D%2B12%5C%20%5Cmu%20C)
Charge of second Particle ![q_2=-8\ \mu C](https://tex.z-dn.net/?f=q_2%3D-8%5C%20%5Cmu%20C)
distance between them ![d=4\ cm](https://tex.z-dn.net/?f=d%3D4%5C%20cm)
![k=9\times 10^{9}](https://tex.z-dn.net/?f=k%3D9%5Ctimes%2010%5E%7B9%7D)
magnetic field due to first charge at mid-way between two charged particles is
![E_1=\frac{kq_1}{r^2}](https://tex.z-dn.net/?f=E_1%3D%5Cfrac%7Bkq_1%7D%7Br%5E2%7D)
![r=\frac{d}{2}=\frac{4}{2}=2\ cm](https://tex.z-dn.net/?f=r%3D%5Cfrac%7Bd%7D%7B2%7D%3D%5Cfrac%7B4%7D%7B2%7D%3D2%5C%20cm)
![E_1=\frac{9\times 10^9\times 12\times 10^{-6}}{(2\times 10^{-2})^2}](https://tex.z-dn.net/?f=E_1%3D%5Cfrac%7B9%5Ctimes%2010%5E9%5Ctimes%2012%5Ctimes%2010%5E%7B-6%7D%7D%7B%282%5Ctimes%2010%5E%7B-2%7D%29%5E2%7D)
(away from it)
Electric field due to ![q_2=-8\ \mu C](https://tex.z-dn.net/?f=q_2%3D-8%5C%20%5Cmu%20C)
![E_2=\frac{kq_2}{r^2}](https://tex.z-dn.net/?f=E_2%3D%5Cfrac%7Bkq_2%7D%7Br%5E2%7D)
![E_2=-\frac{9\times 10^9\times 8\times 10^{-6}}{(2\times 10^{-2})^2}](https://tex.z-dn.net/?f=E_2%3D-%5Cfrac%7B9%5Ctimes%2010%5E9%5Ctimes%208%5Ctimes%2010%5E%7B-6%7D%7D%7B%282%5Ctimes%2010%5E%7B-2%7D%29%5E2%7D)
(towards it)
![E_{net}=E_1+E_2](https://tex.z-dn.net/?f=E_%7Bnet%7D%3DE_1%2BE_2)
(away from first charge)
The answer is 100N. Look up the definition of Newton's third law.