A very long straight current-carrying wire produces a magnetic field of 20 mT at a distance d from the wire. To measure a field
of 5 mT due to this wire, you would have to go to a distance from the wire of
A- 4d.
B - 16d.
C - d sqrt2
D- 8d.
E- 2d.
A- 4d.
B - 16d.
C - d sqrt2
D- 8d.
E- 2d.
1 answer:
Answer:
A. 4d
Explanation:
Magnetic field strength is inversely proportional to distance. So in order to have a smaller magnetic field, we need to move further out from the wire. How far we go exactly can be determined from the formula: B=(μ₀I)/(2πr)
(That is derived from Ampere's Law, which states ∫B•dl=μ₀I)
With that you can set up a ratio between the magnetic fields in both cases. Because the current is the same for both instances, everything reduces out on one side of the equation and leaves you with something that relates the two distances by a ratio of each magnetic field value.
My work is in the attachment, comment for questions.
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The final momentum of the body is equal to 120 Kg.m/s.
<h3>What is momentum?</h3>Momentum can be described as the multiplication of the mass and velocity of an object. Momentum is a vector quantity as it carries magnitude and direction.
If m is an object's mass and v is its velocity then the object's momentum p is: . The S.I. unit of measurement of momentum is kg⋅m/s, which is equivalent to the N.s.
Given the initial momentum of the body = Pi = 20 Kg.m/s
The force acting on the body, Pf = 25 N
The time, Δt = 4-0 = 4s
The Force is equal to the change in momentum: F ×Δt = ΔP
25 × 4 = P - 20
100 = P - 20
P = 100 + 20 = 120 Kg.m/s
Therefore, the final momentum of a body is 120 Kg.m/s.
Learn more about momentum, here:
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Answer:
power emitted is 1.75 W
Explanation:
given data
length l = 5 cm = 5 × m
diameter d = 0.074 cm = 74 × m
total filament emissivity = 0.300
temperature = 3068 K
to find out
power emitted
solution
we find first area that is π×d×L
area = π×d×L
area = π×74 ××5 ×
area = 1162.3892 × m²
so here power emitted is express as
power emitted = E × σ × area × (temperature)^4
put here all value
power emitted = 0.300× 5.67 × 1162.3892 × × (3068)^4
power emitted = 1.75 W