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3 years ago

9

Assume that a lightning bolt can be modeled as a long, straight line of current. If 16 C of charge passes by a point in 1.50 x 1

0-3 s, what is the magnitude of the magnetic field at a perpendicular distance of 27 m from the lighting bolt. Give your answer in LaTeX: \muμT as a whole number.

1 answer:

Reptile [31]3 years ago

7 0

Answer:

$B = 7.9012*10^{-5}T$

Explanation:

To solve the problem, the concepts related to the magnetic field and the current produced in a lightning bolt are necessary.

The current is defined by the load due to time, that is to say

$I= \frac{q}{t}$

Where,

$q= Charge$

$t = time$

So the current can be expressed as:

$I = \frac{16}{1.5*10^{-3}}$

$I = 10666.67A$

Once the current is found it is now possible to find the magnetic field, as this is given by the equation,

$B =\frac{\mu_0}{2\pi}\frac{I}{r}$

Where,

$\mu_0 =$ Permeability Constant

I= Current

r= radius

Replacing the values we have

$B=\frac{4\pi*10^{-7}}{2\pi}(\frac{10666.67}{27})$

$B = 7.9012*10^{-5}T$

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Explanation:

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<h3><u>Answer;</u></h3>

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3 years ago

Draw the following vector quantity Using the coordinate system.

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The given vectors quantities can be described by their properties of both

magnitude and direction.

a. The drawing of the vector extending from point (0, 0) to (190, 0) on the coordinate plane is attached.

b. The velocity vector extending from (0, 0) to (108.76, 50.714) on the coordinate plane is attached.

c. The displacement vector extending from (0, 0) to (30·√2, 30·√2) is attached.

Reasons:

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The direction in which the vector acts = East

Therefore, in vector form, we have;

$\vec{F}$ = 190 × cos(0)·i + 190 × sin(0)·j = 190·i

The vector can be represented by an horizontal line, 190 units long

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The drawing of the vector with the above points using MS Excel is attached.

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Solution;

The vector form of the velocity is; $\vec{v}$ = 120 × cos(25)·i + 120×sin(25)·j, which gives;

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The drawing of the vector is attached

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