Physics, 04.08.2019 00:00 skyyk8107

Two wires carrying equal but opposite currents are twisted together in the construction of a circuit. why does this technique reduce stray magnetic fields?

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The key is to think about it using Integral form of Ampere's Law:
$\oint_c \mathbf{B}.d\mathbf{l}=\mu_0 I_{enc}$

We use this equation like so, create an circular surface in which the wire will go through as shown in the illustration. The current in the wire will also go through the surface. Ampere's law states then that if the current traverses the surface the the current enclosed will be equal to the line integral of the magnetic field H ( $H=\frac{B}{\mu_0}$ in this case) along the closed path that encloses such surface. Due to symmetry the magnetic field will no vary so we end up having a constant magnetic field which we can take out of the integral. We end up having the following:
$\oint_c \mathbf{B}.d\mathbf{l}=B\oint dl=B.2.\pi.r=\mu_0 I_{enc}\impliesB=\frac{\mu_0I}{2\pi r}$
The field will point in the direction given by the right hand rule. We we assumed in the second step using the right hand rule that the field and the vector tangent to the path are parallel. The path is a cirlce so the integral over the path element is the circumference of that cirlcle.
For a parallel just next to the first wire but with current -I will produce a magnetic field Equal to our first in magnitude but with opposite direction, effectively nullifying the first one. Now it way sound bold but twisting the wires together locally means that both wires are always parallel to each other all along them. This implies both magnetic fields will negate as long as the wires are twisted.
Two wires carrying equal but opposite currents are twisted together in the construction of a circuit

Two wires carrying equal but opposite currents are twisted together in the construction of a circuit

Answer from: Quest

firearm identification is the process of analyzing the bullets and cartridge cases left at a crime scene to determine if they came from a particular firearm. marks on the bullets and cartridges may be common to every firearm of that type. these are called class characteristics.

Answer from: Quest

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

$a = 32 ft/s^2$

so as we know that

$a = \frac{v^2}{r}$

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

$a_c = 32 ft/s^2$

$a_t = 0$

now radius of curvature is given as

$r = \frac{v^2}{a_c}$

$r = \frac{150^2}{32}$

$r = 703.125ft$

so the radius of curvature just after it is dropped is 703.125 ft

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