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Anika [276]
3 years ago
9

A wire 25 m long carries a current of 12 A from west to east. If the magnetic force on the wire due to Earth’s magnetic field is

upward (away from Earth) and has a magnitude of 4.0 × 10⁻² N, find the magnitude and direction of the magnetic field at this location.​
Physics
1 answer:
ra1l [238]3 years ago
6 0
<h2>Answer:</h2>

<em>1.33 x 10⁻ ⁴ T outwards.</em>

<em></em>

<h2>Explanation:</h2>

The equation for the magnetic force (F) on a wire whose length is L and carrying a current I in a magnetic field (B) that is uniform is given by;

F = ILB sin θ          ---------------------(i)

Where;

θ = angle between the direction of the current and that of the magnetic field.

From the question,

F = 4.0 × 10⁻² N

I = 12A

L = 25m

θ = 90°

<em>Substitute these values into equation(i) and solve as follows;</em>

4.0 × 10⁻² = 12 x 25 x B x sin 90°

4.0 × 10⁻² = 300 x B x 1

4.0 × 10⁻² = 300B

0.04 = 300B

B = \frac{0.04}{300}

B = 0.000133

B = 1.33 x 10⁻ ⁴ T

To get the direction of the magnetic field, the right-hand rule is used.

If the right hand fingers are positioned in the correct order specified by the right hand rule, then it would be seen that the magnetic field is directed outwards.

Therefore, the magnitude and direction of the magnetic field at this location is <em>1.33 x 10⁻ ⁴ T outwards.</em>

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Consider the points below. P(1, 0, 1), Q(−2, 1, 4), R(6, 2, 7) (a) Find a nonzero vector orthogonal to the plane through the poi
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Answer:

a) (0, -33, 12)

b) area of the triangle : 17.55 units of area

Explanation:

<h2>a) </h2>

We know that the cross product of linearly independent vectors \vec{A} and \vec{B} gives us a nonzero, orthogonal to both, vector. So, if we can find two linearly independent vectors on the plane through the points P, Q, and R, we can use the cross product to obtain the answer to point a.

Luckily for us, we know that vectors \vec{A} = \vec{P}-\vec{Q} and \vec{B} = \vec{R} - \vec{Q} are living in the plane through the points P, Q, and R, and are linearly independent.

We know that they are linearly independent, cause to have one, and only one, plane through points P Q and R, this points must be linearly independent (as the dimension of a plane subspace is 3).

If they weren't linearly independent, we will obtain vector zero as the result of the cross product.

So, for our problem:

\vec{A} = \vec{P} - \vec{Q} \\\\\vec{A} = (1,0,1) - (-2,1,4)\\\\\vec{A} = (1 +2,0-1,1-4)\\\\\vec{A} = (3,-1,-3)

\vec{B} = \vec{R} - \vec{Q} \\\\\vec{B} = (6,2,7) - (-2,1,4)\\\\\vec{B} = (6 +2,2-1,7-4)\\\\\vec{B} = (8,1,3)

\vec{A} \times  \vec{B} = (A_y B_z - B_y A_z) \  \hat{i} - ( A_x B_z-B_xA_z) \ \hat{j} + (A_x B_y - B_x A_y ) \ \hat{k}

\vec{A} \times  \vec{B} = ( (-1) * 3 - 1 * (-3) ) \  \hat{i} - ( 3 * 3 - 8 * (-3)) \ \hat{j} + (3 * 1 - 8 * (-1) ) \ \hat{k}

\vec{A} \times  \vec{B} = ( - 3 + 3 ) \  \hat{i} - ( 9 + 24 ) \ \hat{j} + (3 + 8 ) \ \hat{k}

\vec{A} \times  \vec{B} = 0 \  \hat{i} - 33 \ \hat{j} + 12 \ \hat{k}

\vec{A} \times  \vec{B} =(0, -33, 12)

<h2>B)</h2>

We know that \vec{A} and \vec{B} are two sides of the triangle, and we also know that we can use the magnitude of the cross product to find the area of the triangle:

|\vec{A} \times  \vec{B} | = 2 * area_{triangle}

so:

\sqrt{(-33)^2 + (12)^2} = 2 * area_{triangle}

\sqrt{1233} = 2 * area_{triangle}

35.114= 2 * area_{triangle}

17.55 \ units \  of \ area =  area_{triangle}

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3 years ago
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