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Tamiku [17]
4 years ago
14

If the direction of a magnetic field around a vertical electrical wire is counter clockwise, what direction is the electrical cu

rrent moving?
From the top of the wire to the bottom
From the bottom of the wire to the top
Counter clockwise
Clockwise
Physics
1 answer:
strojnjashka [21]4 years ago
8 0

Direction of the current:

From the bottom of the wire to the top

Explanation:

The magnetic field produced by a current-carrying wire forms concentric circles around the wire; its magnitude is given by

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

whrere

\mu_0 is the vacuum permeability

I is the current in the wire

r is the distance from the wire

The direction of the magnetic field produced by a current-carrying wire can be found by using the right-hand rule:

- The thumb of the right hand is placed in the same direction as the current

- The other fingers, wrapped around the thumb, give the direction of the magnetic field

In this case, the wire is vertical and the magnetic field is counter clockwise (looking from the top), therefore we need to put the thumb from the bottom to the top in order to get the same direction for the field: therefore, the direction of the current is

From the bottom of the wire to the top

Learn more about magnetic fields:

brainly.com/question/3874443

brainly.com/question/4240735

#LearnwithBrainly

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If =12a andthe distance from each wire to point p is 0.12m, then what is the magnitude of the magnetic force per unit length on
vaieri [72.5K]

The magnitude of the magnetic force per unit length on the top wire is

2×10⁻⁵  N/m

<h3>How can we calculate the magnitude of the magnetic force per unit length on the top wire ?</h3>

To calculate the magnitude of the magnetic force per unit length on the top wire, we are using the formula

F= \frac{\mu_0 I_f}{2\pi d}

Here we are given,

\mu_0= magnetic permeability

= 4\pi×10⁻⁷ H m⁻¹

If= 12 A

d= distance from each wire to point.

=0.12m

Now we put the known values in the above equation, we get

F= \frac{\mu_0 I_f}{2\pi d}

Or, F = \frac{4\pi \times 10^{-7}\times  12}{2\pi \times 0.12}

Or, F= 2×10⁻⁵ N/m.

From the above calculation, we can conclude that the magnitude of the magnetic force per unit length on the top wire is 2×10⁻⁵ N/m.

Learn more about magnetic force:

brainly.com/question/2279150

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7 0
2 years ago
A person exerts a horizontal force of 190 n in the test apparatus shown in the figure. Find the horizontal force that his flexor
DochEvi [55]

Answer:

Check explanation.

Explanation:

From the question, we know that The person exerted 190N, force on the flexor is unknown. Since, we don't have access to our diagram, we have to make one or two assumption; (1) that ba= 0.3 m, ac= 0.05.

Therefore, the horizontal force that his flexor muscle exerts on his forearm,F(flexor) = (190N) × (0.3m) / 0.05m.

The horizontal force that his flexor muscle exerts on his forearm,F(flexor) = 57 Nm/ 0.05 m.

The horizontal force that his flexor muscle exerts on his forearm,F(flexor) = 1140N.

5 0
3 years ago
What is the wavelength of this wave?<br><br> 1<br> 2<br> 3<br> 4
Mama L [17]
2 would be your answer
3 0
3 years ago
Read 2 more answers
A blow-dryer and a vacuum cleaner each operate with a voltage of 120 V. The current rating of the blow-dryer is 10 A, and that o
Katyanochek1 [597]

Answer:

Part a)

P = 1200 Watt

Part b)

P = 480 W

Part c)

\frac{E_{bd}}{E_{vc}} = 5 : 2

Explanation:

Part a)

Power consumed by blow dryer is given as

P = iV

here supply voltage is given as

V = 120 volts

current rating of blow dryer is given as

i = 10 A

P = (120 V)(10 A)

P = 1200 W

Part b)

Power consumed by vacuum cleaner is given as

P = iV

here supply voltage is given as

V = 120 volts

current rating of vacuum cleaner is given as

i = 4 A

P = (120 V)(4 A)

P = 480 W

Part c)

Energy consumed in a given interval of time is given as

Energy = power \times time

now we have to find the ratio of energy consumed in time interval of 11 minutes

so we have

\frac{E_{bd}}{E_{vc}} = \frac{P_{bd} t}{P_{vc} t}

now we have

\frac{E_{bd}}{E_{vc}} = \frac{1200 W\times 11 min}{480 W \times 11 min}

\frac{E_{bd}}{E_{vc}} = 5 : 2

5 0
4 years ago
A wave travels at a frequency of 387 Hz. What is<br>the period of the wave?<br>​
Ulleksa [173]

Answer:

¹/₃₈₇ second

Explanation:

<em>The period of a wave is the reciprocal of its frequency.</em>

So, simply, the frequency is ¹/₃₈₇ second(s), as that is the reciprocal of the frequency, 387 Hz.

5 0
4 years ago
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