Answer:
Rate of change of magnetic field is
Explanation:
We have given diameter of the circular loop is 13 cm = 0.13 m
So radius of the circular loop 
Length of the circular loop 
Wire is made up of diameter of 2.6 mm
So radius 
Cross sectional area of wire 
Resistivity of wire 
Resistance of wire 
Current is given i = 11 A
So emf 
Emf induced in the coil is 


A plane mirror always forms a virtual image. the image and the object are the same distance from a flat mirror, the image size is the same as the object, and the image is upright!
Given that,
At 3.00 m from a source that is emitting sound uniformly in all directions, the sound level is 60.0 dB.
To find,
The distance from the source would the sound level be one-fourth the sound level at 3.00 m.
Solution,
The intensity from a source is inversely proportional to the distance.
Let I₁ = 60 dB, r₁ = 3 m, I₂ = 60/4 = 15 dB, r₂ =?
Using relation :

So, at a distance of 6 m the sound level will be one fourth of the sound level at 3 m.