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miss Akunina [59]

3 years ago

14

Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter of 23.0 and carries a cl

ockwise current of 20.0 , as viewed from above, and the outer wire has a diameter of 38.0. What must be the direction (as viewed from above) of the current in the outer wire so that the net magnetic field due to this combination of wires is zero at the common center of the wires? What must be the magnitude of the current in the outer wire so that the net magnetic field due to this combination of wires is zero at the common center of the wires?

1 answer:

Bond [772]3 years ago

7 0

Answer:

I = 33.04A

Explanation:

The magnetic field at the center of a loop of wire is proportional to the current flowing through the wire and inversely proportional to the radius

So therefore we have:

20/23 = I/38.0

We then solve for I

cross multiplying we have 20 x 38 = 23 x I

I = 20 x 38/ 23

I = 33.04A

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A point charge of 5.7 µc moves at 4.5 × 105 m/s in a magnetic field that has a field strength of 3.2 mt, as shown in the diagram

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The magnitude of the force on the charge by the influence of the magnetic field will be 6.6*10^-3 N

<h3>What is magnetic force?</h3>

Magnetic force, attraction or repulsion that arises between electrically charged particles because of their motion.The magnitude of the magnetic force acting on the charge is given by:

$\rm F=qvBsin\theta$

where

The magnitude of the charge $q=5.7\ \mu C =5.7\times 10^{-6} \ C$

The velocity of the charge $v=4.5\times 10^5\ \frac{m}{s}$

The magnitude of the magnetic field $3.2\ mT=0.0032\ T$

The angle between the directions of v and B $\theta =90^o-37^o=53^o$

By substituting the values we will get:

$F=(5.7\times 10^}-6})(4.5\times 10^5)(0.0032)(Sin53^o)$

$F=6.6\times 10^{-3}\ N$

Hence the magnitude of the force on the charge by the influence of the magnetic field will be 6.6*10^-3 N

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brainly.com/question/14411049

3 0

2 years ago

An object is thrown directly downward from the top of a very tall building. The speed of the object just as it is released is 17

Softa [21]

Answer:

distance cover is = 102.53 m

Explanation:

Given data:

speed of object is 17.1 m/s

$t_1 = 3.32 sec$

$t_2 = 5.08 sec$

from equation of motion we know that

$d_1 = vt_1 + \frac{1}{2} gt_1^2$

where d_1 is distance covered in time t1

so $d_1 = 17.1 \times 3.32 + \frac{1}{2} 9.8 \times 3.32^2$ =

$d_1 = 110.78 m$

$d_2 = vt_2 + \frac{1}{2} gt_2^2$

where d_2 is distance covered in time t2

$d_2 = 17.1 \times 5.08 + \frac{1}{2}\times9.8 \times 5.08^2$

$d_2 = 213.31 m$

distance cover is = 213.31 - 110.78 = 102.53 m

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3 years ago

In one year, a large tree can remove from the air the same amount of carbon dioxide produced by a car travelling 500 miles. If N

marissa [1.9K]

I am so sure it's 600,000 x 500 so you get 300,000,000

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3 years ago

A student grabs a piece of paper and tears it from the middle. Which statement best explains the forces involved?

PolarNik [594]

Answer:

b. The normal force between the molecules of the paper is overcome by the contact force of the hands.

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2 years ago

Read 2 more answers

Which temperature is the hottest? 98 F or 39 C or 303K?<br> F= 1.8C + 32<br> C= (F-32)/1.8

sergejj [24]

Answer:

The hottest temperature is $T_2 = 39^o C$

Explanation:

From the question we are given

$T_1 = 98 F$

$T_2 = 39^oC$

$T_3 = 303 \ K$

Generally converting $T_3$ to Fahrenheit

$T_3' = (T_3 -273 ) * \frac{9}{5} + 32$

=> $T_3' = (303 -273 ) * \frac{9}{5} + 32$

=> $T_3' = 86 F$

Converting $T_2$ to Fahrenheit

$T_2' = T_2 * \frac{9}{5} + 32$

=> $T_2' = 39 * \frac{9}{5} + 32$

=> $T_2' =102.2 F$

Now comparing the temperature in Fahrenheit we see that $T_2$ is the hottest

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3 years ago