A man drops a penny V=0 off the top of the Golden Gate Bridge how fast will the penny be moving when it hits the ground? The gol
1 answer:
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Answer:
The current flowing through the outer coils is
Explanation:
From the question we are told that
The number of turn of inner coil is
The radius of inner coil is
The current flowing through the inner coil is
The number of turn of outer coil is
The radius of outer coil is
For net magnetic field at the common center of the two coils to be zero the current flowing in the outer coil must be opposite to current flowing inner coil
The magnetic field due to inner coils is mathematically represented as
The magnetic field due to inner coils is mathematically represented as
Now for magnetic field at center to be zero
So
=>
Answer:
the decay of half of the nuclei only a half-life has passed , b) in rock time it is 1 108 years
Explanation:
The radioactive decay is given by
N = N₀
If half of the atoms have decayed
½ N₀ = N₀
½ =
₀
Ln 0.5 = - λ t
t = - ln 0.5 /λ
The definition of average life time is
= ln 2 / λ
λ = ln 2 /
λ = 0.693 / 100 10⁶
λ = 0.693 10⁻⁸ years
We replace
t = -ln 0.5 / 0.693 10⁻⁸
t = 10⁸ years
We see that for the decay of half of the nuclei only a half-life has passed
b) in rock time it is 1 108 years
Answer:
Explanation:
we can consider an element of radius r < a and thickness dr. and Area of this element is
since current density is given
then , current through this element will be,
integrating on both sides between the appropriate limits,
Magnetic field can be found by using Ampere's law
for points inside the wire ( r<a)
now, consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius r.
by applying the Ampere's law, we can write
by symmetry will be of uniform magnitude on this loop and it's direction will be tangential to the loop.
Hence,
now using equation 1, putting the value of k,
B)
now, for points outside the wire ( r>a)
consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius l.
applying the Ampere's law
by symmetry will be of uniform magnitude on this loop and it's direction will be tangential to the loop. Hence
again using,equaiton 1,