A magnetic field is uniform over a flat, horizontal circular region with a radius of 1.50 mm, and the field varies with time. In
itially the field is zero and then changes to 1.50 T, pointing upward when viewed from above, perpendicular to the circular plane, in a time of 125 ms what is the average induced emf around the border of the circular region?
The average induced emf around the border of the circular region is .
Explanation:
Given that,
Radius of circular region, r = 1.5 mm
Initial magnetic field, B = 0
Final magnetic field, B' = 1.5 T
The magnetic field is pointing upward when viewed from above, perpendicular to the circular plane in a time of 125 ms. We need to find the average induced emf around the border of the circular region. It is given by the rate of change of magnetic flux as :
So, the average induced emf around the border of the circular region is .
Given there are three blocks of masses , and (ref image in attachment)
When all three masses move together at an acceleration a, the force F is given by
F = ( + + ) *a ................(equation 1)
Also it is given that does not move with respect to , which gives tension T is exerted on pulley by only, Hence tension T is
T = *a ..........(equation 2)
There is also also tension exerted by . There are two components here: horizontal due to acceleration a and vertical component due to gravity g. Thus tension is given by
T = ................(equation 3)
From equation 2 and 3, we get
*a =
Squaring both sides we get
* = * (+)
* = ( * )+ ( *)
( - ) * = *
= */( - )
Taking square root on both sides, we get acceleration a
a = *g/()
Hence substituting the value of a in equation 1, we get
Explanation:The magnetic pole near earth's geographic north pole is actually the south magnetic pole. When it comes to magnets, opposites attract. This fact means that the north end of a magnet in a compass is attracted to the south magnetic pole, which lies close to the geographic north pole.