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masya89 [10]
3 years ago
11

A wire loop is suspended from a string that is attached to point P in the drawing. When released, the loop swings downward, from

left to right, through a uniform magnetic field, with the plane of the loop remaining perpendicular to the plane of the paper at all times. Determine the direction of the current induced in the loop as it swings past the locations labeled (a) I and (b) II. Specify the direction of the current in terms of the points x, y, and z on the loop (e.g., x→y→z or z→y→x). The points x, y, and z lie behind the plane of the paper. What is the direction of the induced current at the locations (c) I and (d) II when the loop swings back, from right to left?

Physics
1 answer:
Naddik [55]3 years ago
5 0

Complete Question

The complete question iws shown on the first uploaded image  

Answer:

a

    y \to z \to x

b

  x \to z \to y

Explanation:

Now looking at the diagram let take that the magnetic field is moving in the x-axis

 Now the magnetic force is mathematically represented as

             F =  I L x B

Note (The x is showing cross product )

Note the force(y-axis) is perpendicular to the field direction (x-axis)

Now when the loop is swinging forward

 The motion of the loop is  from   y to z to to x to y

Now since the force is perpendicular to the motion(velocity) of the loop

Hence the force would be from z to y and back to z  

and from lenze law the induce current opposes the force so the direction will be from y to z to x

Now when the loop is swinging backward

   The motion of the induced current will now be   x to z to y

 

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Answer:

Explanation:

Given

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A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out
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Answer:

The horizontal distance is 4.823 m

Solution:

As per the question:

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Now,

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Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

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Initially, the man starts, form rest thus the velocity at start will be zero and hence the initial Kinetic energy will also be zero.

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Making use of the principle of conservation of momentum in order to calculate the velocity after the inclusion, v' of the backpack:

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t = \sqrt{\frac{2h}{g}}

t = \sqrt{\frac{2\times 2}{9.8} = 0.638\ s

Now, the horizontal distance is given by:

x = v't = 7.56\times 0.638 = 4.823\ m

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