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abruzzese [7]
3 years ago
15

Refer to a long, straight wire carrying constant current I. What can be concluded about the magnitude of the magnetic field at d

istance a from the wire?
Physics
1 answer:
sergij07 [2.7K]3 years ago
5 0

Answer:

<em>"the magnitude of the magnetic field at a point of distance a around a wire, carrying a constant current I, is inversely proportional to the distance a of the wire from that point"</em>

Explanation:

The magnitude of the magnetic field from a long straight wire (A approximately a finite length of wire at least for close points around the wire.) decreases with distance from the wire. It does not follow the inverse square rule as is the electric field from a point charge. We can then say that<em> "the magnitude of the magnetic field at a point of distance a around a wire, carrying a constant current I, is inversely proportional to the distance a of the wire from that point"</em>

From the Biot-Savart rule,

B = μI/2πR

where B is the magnitude of the magnetic field

I is the current through the wire

μ is the permeability of free space or vacuum

R is the distance between the point and the wire, in this case is = a

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PART A)

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n_1sin i = n_2 sin r

here we know that

n_1 = 1.5

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1.5 sin25 = 1 sin r

r = 39.3 degree

so it will refract by angle 39.3 degree

PART B)

Here as we can see that image formed on the other side of lens

So it is a real and inverted image

Also we can see  that size of image is lesser than the size of object here

Here we can use concave mirror to form same type of real and inverted image

PART C)

As per the mirror formula we know that

\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}

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A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in
koban [17]

Answer:

a) The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

b) The electric force is 2.0\times 10^{-6} newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

E = \frac{F_{e}}{q} (1)

Where:

E - Electric field, measured in newtons per coulomb.

F_{e} - Electric force, measured in newtons.

q - Electric charge, measured in coulombs.

If we know that F_{e} = 4.0\times 10^{-6}\,N and q = 2.0\times 10^{-8}\,C, then the electric field at that point is:

E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}

E = 2.0\times 10^{2}\,\frac{N}{C}

The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

b) If we know that E = 2.0\times 10^{2}\,\frac{N}{C} and q = 1.0\times 10^{-8}\,C, then the electric force is:

F_{e} = E\cdot q

F_{e} = \left(2.0\times 10^{2}\,\frac{N}{C} \right)\cdot (1.0\times 10^{-8}\,C)

F_{e} = 2.0\times 10^{-6}\,N

The electric force is 2.0\times 10^{-6} newtons.

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Which vector best represents the force that could act concurrently with force A to produce force B
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