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likoan [24]
3 years ago
11

You hold a wire coil so that the plane of the coil is perpendicular to a magnetic field b⃗ .part aif the magnitude of b⃗ increas

es while its direction remains unchanged, how will the magnetic flux through the coil change?
Physics
2 answers:
JulsSmile [24]3 years ago
4 0

The magnetic flux linked with the coil will <u>increase because the magnitude of the magnetic field is increased.</u>

Further Explanation:

The magnetic flux linked with a coil placed perpendicular to the direction of magnetic field is given as:

\boxed{\phi =BA\cos\theta }

Here, B is the magnetic field, A is the cross-sectional and \theta is the angle between the coil and the magnetic field.

The above expression shows that the magnetic flux of a coil is directly proportional to the magnetic field lines passing through the coil because more the strength of the field more will be the number of magnetic field lines passing through the coil.

The magnetic flux induced in the coil is directly proportional to the area of the coil because due to larger area of the coil more number of magnetic field lines will pass through it and the change in position of the coil will also change the magnetic flux linked with the coil.

Thus, in the above case, if the magnetic field is increased in the region, the magnetic flux of the coil will also increase.

So, the magnetic flux linked with the coil will<u> increase because the magnitude of the magnetic field is increased. </u>

<u> </u>

Learn More:

1. A 16.2 g piece of aluminum (which has a molar heat capacity of 24.03 j/°c•mol) is heated to 82.4°c brainly.com/question/7175743

2. What evidence is there that electrons move around in definite pathways around the nucleus brainly.com/question/2927304

Answer Details:

Grade: Senior School

Subject: Physics

Chapter: Electromagnetic Induction

Keywords:

Magnetic flux, magnetic field, linked with the coil, area of cross-section of coil, flux increases, increased magnetic field, and direction remains unchanged.

postnew [5]3 years ago
3 0
Magnetic flux Ф is given by
Ф = ∫В.dA

dA is an element of area of the surface, the integration is carried over the entire surface through which we wish to calculate the flux. B is magnetic field.

If direction of coil is remained the same, area of the coil remains the same. But as we increase the magnetic field, the number of field lines passing through the coil increases. Therefore the magnetic flux will increase.
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Answer:

a) F_{r}= -583.72MN i + 183.47MN j + 6.05GN k

b) E=3.04 \frac{GN}{C}

Step-by-step explanation.

In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).

Once we drew the points, we can start calculating the forces:

r_{AP}^{2}=(3-0)^{2}+(3-0)^{2}+(5+0)^{2}

which yields:

r_{AP}^{2}= 43 m^{2}

(I will assume the positions are in meters)

Next, we can make use of the force formula:

F=k_{e}\frac{q_{1}q_{2}}{r^{2}}

so we substitute the values:

F_{AP}=(8.99x10^{9})\frac{(1C)(2C)}{43m^{2}}

which yields:

F_{AP}=418.14 MN

Now we can find its components:

F_{APx}=418.14 MN*\frac{3}{\sqrt{43}}i

F_{APx}=191.30 MNi

F_{APy}=418.14 MN*\frac{3}{\sqrt{43}}j

F_{APy}=191.30MN j

F_{APz}=418.14 MN*\frac{5}{\sqrt{43}}k

F_{APz}=318.83 MN k

And we can now write them together for the first force, so we get:

F_{AP}=(191.30i+191.30j+318.83k)MN

We continue with the next force. The procedure is the same so we get:

r_{BP}^{2}=(3-1)^{2}+(3-1)^{2}+(5+0)^{2}

which yields:

r_{BP}^{2}= 33 m^{2}

Next, we can make use of the force formula:

F_{BP}=(8.99x10^{9})\frac{(4C)(2C)}{33m^{2}}

which yields:

F_{BP}=2.18 GN

Now we can find its components:

F_{BPx}=2.18 GN*\frac{2}{\sqrt{33}}i

F_{BPx}=758.98 MNi

F_{BPy}=2.18 GN*\frac{2}{\sqrt{33}}j

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F_{BPz}=2.18 GN*\frac{5}{\sqrt{33}}k

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And we can now write them together for the second, so we get:

F_{BP}=(758.98i + 758.98j + 1897k)MN

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F_{CP}=(8.99x10^{9})\frac{(7C)(2C)}{30m^{2}}

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F_{CPx}=4.20 GN*\frac{-2}{\sqrt{30}}i

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F_{CPy}=4.20 GN*\frac{2}{\sqrt{30}}j

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And we can now write them together for the third force, so we get:

F_{CP}=(-1.534i - 0.76681j +3.83k)GN

So in order to find the resultant force, we need to add the forces together:

F_{r}=F_{AP}+F_{BP}+F_{CP}

so we get:

F_{r}=(191.30i+191.30j+318.83k)MN + (758.98i + 758.98j + 1897k)MN + (-1.534i - 0.76681j +3.83k)GN

So when adding the problem together we get that:

F_{r}=(-0.583.72i + 0.18347j +6.05k)GN

which is the answer to part a), now let's take a look at part b).

b)

Basically, we need to find the magnitude of the force and divide it into the test charge, so we get:

F_{r}=\sqrt{(-0.583.72)^{2} + (0.18347)^{2} +(6.05)^{2}}

which yields:

F_{r}=6.08 GN

and now we take the formula for the electric field which is:

E=\frac{F_{r}}{q}

so we go ahead and substitute:

E=\frac{6.08GN}{2C}

E=3.04\frac{GN}{C}

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