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Answer:
Explanation:
The image is real light rays actually focus at the image location). As the object moves towards the mirror the image location moves further away from the mirror and the image size grows (but the image is still inverted).
Answer:
The diameter of wire should be
m
Explanation:
Given:
Current density
![\frac{A}{m^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7BA%7D%7Bm%5E%7B2%7D%20%7D)
Current
A
From the formula of current density,
![J = \frac{I}{A}](https://tex.z-dn.net/?f=J%20%3D%20%5Cfrac%7BI%7D%7BA%7D)
Where
area of cylindrical wire = ![\pi r^{2}](https://tex.z-dn.net/?f=%5Cpi%20r%5E%7B2%7D)
![\pi r^{2} = \frac{I}{J}](https://tex.z-dn.net/?f=%5Cpi%20r%5E%7B2%7D%20%3D%20%5Cfrac%7BI%7D%7BJ%7D)
![r^{2} = \frac{I}{\pi J }](https://tex.z-dn.net/?f=r%5E%7B2%7D%20%3D%20%5Cfrac%7BI%7D%7B%5Cpi%20J%20%7D)
![r = \sqrt{\frac{0.64}{3.14 \times 500 \times 10^{4} } }](https://tex.z-dn.net/?f=r%20%3D%20%5Csqrt%7B%5Cfrac%7B0.64%7D%7B3.14%20%5Ctimes%20500%20%5Ctimes%2010%5E%7B4%7D%20%7D%20%7D)
m
For finding the diameter of wire,
![d = 2r](https://tex.z-dn.net/?f=d%20%3D%202r)
m
Therefore, the diameter of wire should be
m