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ehidna [41]
3 years ago
8

Two insulated wires, each 2.50 m long, are taped together to form a two-wire unit that is 2.50 m long. One wire carries a curren

t of 7.00 A; the other carries a smaller current I in the opposite direction. The two wire unit is placed at an angle of 79.0° relative to a magnetic field whose magnitude is 0.360 T. The magnitude of the net magnetic force experienced by the two-wire unit is 3.13 N. What is the current I?
Physics
1 answer:
liubo4ka [24]3 years ago
3 0

Answer:

3.46 A

Explanation:

The force (F) exerted on a wire of a particular length (L) carrying current (I) through a magnetic field (B) at an angle (θ) to the magnetic field is given as

F = (B)(I)(L) sin θ

F = 3.13 N

B = 0.360 T

I = ?

L = 2.50 m

θ = 79°

3.13 = (0.360 × I × 2.5 × sin 79°)

0.8835 I = 3.13

I = 3.54 A

But this is the resultant current in this magnetic field.

Since the two wires are conducting current in opposite directions,

Resultant current = 7 - (current in the other wire)

Current in the other wire = 7 - 3.54 = 3.46 A

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3 years ago
Suppose you walk 11 m in a direction exactly 24° south of west then you walk 21 m in a direction exactly 39° west of north. 1) H
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Answer:

(1) 42.94 m

(2) 16.02^\circ

Explanation:

Let us first draw a figure, for the given question as below:

In the figure, we assume that the person starts walking from point A to travel 11 m exactly 24^\circ south of west to point B and from there, it walks 21 m exactly 39^\circ west of north to reach point C.

Let us first write the two displacements in the vector form:

\vec{AB} = (-11\cos 24^\circ\ \hat{i}-11\sin 24^\circ\ \hat{j})\ m =(-10.05\ \hat{i}-4.47\ \hat{j})\ m\\\vec{BC} = (-21\sin 39^\circ\ \hat{i}+21\cos 39^\circ\ \hat{j})\ m =(-31.22\ \hat{i}+16.32\ \hat{j})\ m

Now, the vector sum of both these vectors will give us displacement vector from point A to point C.

\vec{AC}=\vec{AB}+\vec{BC}\\\Rightarrow \vec{AC}=(-10.05\ \hat{i}-4.47\ \hat{j})\ m+(-31.22\ \hat{i}+16.32\ \hat{j})\ m\\\Rightarrow \vec{AC}=(-41.25\ \hat{i}+11.85\ \hat{j})\ m

Part (1):

the magnitude of the shortest displacement from the starting point A to point the final position C is given by:

AC=\sqrt{(-41.25)^2+(11.85)^2}\ m= 42.94\ m

Part (2):

As the vector AC is coordinates lie in the third quadrant of the cartesian vector plane whose angle with the west will be positive in the north direction.

The angle of the shortest line connecting the starting point and the final position measured north of west is given by:

\theta = \tan^{-1}(\dfrac{11.85}{41.27})\\\Rightarrow \theta = 16.02^\circ

4 0
3 years ago
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A bulldozer does 4,500 J of work to push a mound of soil to the top of a ramp that is 15 m high. The ramp is at an
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Answer:

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Plugging in the values and solving for F:

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