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Ray Of Light [21]
2 years ago
15

As viewed from above in this picture, what direction will the current be in the coil of wire that will cause the loop to rotate

in the manner shown?
There is not enough information to tell
Towards the top right corner of the picture
Towards the bottom left corner of the picture
Counterclockwise

Physics
1 answer:
Gala2k [10]2 years ago
7 0

Answer:

When viewed from above, the current in the coil should point towards the top-right corner of the picture.

Explanation:

The current in this coil have only two possible directions: clockwise or counter-clockwise. However, since the diagram shows the coil from above, not from a cross-section, just saying clockwise or counter-clockwise might be ambiguous. The statement that the current is directed towards the top-right corner of the picture is equivalent to saying that when viewed from the lower-right corner of this diagram, the current in the coil is moving clockwise.

Note that at the center of this picture, the current is parallel to the magnetic field- there will be no force on the coil at that position. On the other hand, (also when viewed from above,) at the top-right corner and the lower-left corner of the coil, the current in the coil will be perpendicular to the magnetic field. That's where the force on the coil will be the strongest.

With that in mind, apply the right-hand rule to find the direction of the force on the coil in each of the two possibilities.

Assume that when viewed from above, the current is flowing towards the top-right corner of the picture. Consider the wire near the top-right corner of this coil (as viewed above on this picture.) The current will be going into the picture into the magnetic field. By the right-hand rule, the current on the wire near that point should be pointing towards the bottom of this picture. (Point fingers on the right hand in the direction of the current I. Rotate the right hand such that when curling the fingers, they point in the direction of the magnetic field B. The direction of the right thumb should now point in the direction of the force on the wire F.)

Based on the same assumption, the current in the wires near the bottom left corner of this coil will be pointing out of the picture. By the right hand rule, the magnetic force on the coil in that region should be pointing towards the top of this picture. Combing these two forces, the coil would indeed be rotating around the center of this picture in the direction shown in the diagram.

It can also be shown that if the current points towards the bottom left corner of the picture when viewed from above, the coil will be rotating about the center of this picture in the opposite direction.

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A uniformly charged ring of radius 10.0 cm has a total charge of 50.0 μC Find the electric field on the axis of the ring at 30.0
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Answer: 4.27 *10^6 N/C

Explanation: In order to calculate the electric field along the axis of charged ring we have to use the following expression:

E=k*x/(a^2+x^2)^3/2    where a is the ring radius and x the distance to the point measured from the center of the ring.

Replacing the data we have:

E= (9* 10^9* 0.3* 50 * 10^-6)/(0.1^2+0.3^2)^3/2

then

E=4.27 * 10^6 N/C

8 0
3 years ago
At the local playground, a 21-kg child sits on the right end of a horizontal teeter-totter, 1.8 m from the pivot point. On the l
Ksju [112]

Answer:

By convention a negative torque leads to clockwise rotation and a positive torque leads to counterclockwise rotation.

here weight of the child =21kgx9.8m/s2 = 205.8N

the torque exerted by the child Tc = - (1.8)(205.8) = -370.44N-m ,negative sign is inserted because this torque is clockwise and is therefore negative by convention.

torque exerted by adult Ta = 3(151) = 453N , counterclockwise torque.

net torque Tnet = -370.44+453 =82.56N , which is positive means counterclockwise rotation.

b) Ta = 2.5x151 = 377.5N-m

Tnet = -370.44+377.5 = 7.06N-m , positive ,counterclockwise rotation.

c)Ta = 2x151 = 302N-m

Tnet = -370.44+302 = -68.44N-m, negative,clockwise rotation.

5 0
2 years ago
A 65-kg woman in an elevator is accelerating upward at a rate of 0.6 m/s2. The gravitational force is ___ N?
Nastasia [14]

Answer:

The gravitational force is 130.

Explanation:

During this problem you have to multiply the 65 and the 0.6.

5 0
2 years ago
an object traveling 200 feet per second slows to 50 feet per second in 5 seconds. Calculate the acceleration of the object
emmainna [20.7K]

Answer:

The acceleration of the object is -30\ m/s^2

Explanation:

Given:

Initial velocity of object v_i = 200 feet/second

Final velocity of object v_f = 50 feet/second

Time of travel = 5 seconds

To calculate acceleration of the object we will find the rate of change of velocity with respect to time.

So, acceleration a is given by:

a=\frac{v_f-v_i}{t}

where v_f represents final velocity, v_i represents initial velocity and t is time of travel.

Plugging in values to evaluate acceleration.

a=\frac{50-200}{5}

a=\frac{-150}{5}

a=-30\ m/s^2

The acceleration of the object is -30\ m/s^2 (Answer). The negative sign shows the object is slowing down.

4 0
3 years ago
An object of mass 300 g, moving with an initial velocity of 5.00i-3.20j m/s, collides with an sticks to an object of mass 400 g,
Alexus [3.1K]

Answer:

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

Explanation:

Mass of object 1 , m₁ = 300 g = 0.3 kg

Mass of object 2 , m₂ = 400 g = 0.4 kg

Initial velocity of object 1 , v₁ = 5.00i-3.20j m/s

Initial velocity of object 2 , v₂ = 3.00j m/s

Mass of composite = 0.7 kg

We need to find final velocity of composite.

Here momentum is conserved.

Initial momentum = Final momentum

Initial momentum = 0.3 x (5.00i-3.20j) + 0.4 x 3.00j = 1.5 i + 0.24 j kgm/s

Final momentum = 0.7 x v = 0.7v kgm/s

Comparing

1.5 i + 0.24 j = 0.7v

v = 2.14 i + 0.34 j

Magnitude of velocity      

       v=\sqrt{2.14^2+0.34^2}=2.17m/s

Direction,  

       \theta =tan^{-1}\left ( \frac{0.34}{2.14}\right )=9.03^0

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

7 0
3 years ago
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