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Vadim26 [7]
3 years ago
10

A square loop of wire with a small resistance is moved with constant speed from a field free region into a region of uniform B f

ield (B is constant in time) and then back into a field free region to the right. The self inductance of the loop is negligible
True/False:

a. When leaving the field the coil experiences a magnetic force to the left.

b. Upon entering the field, a clockwise current flows in the loop.

c. Upon leaving the field, a clockwise current flows in the loop.

d. When entering the field the coil experiences a magnetic force to the right.
Physics
1 answer:
Misha Larkins [42]3 years ago
8 0

Answer:

A) False

B) False

C) True

D) False

Explanation:

A) False, because when leaving the field, the coil experiences a magnetic force to the right.

B) When the loop is entering the field, the magnetic flux through it will increase. Thus, induced magnetic field will try to decrease the magnetic flux i.e. the induced magnetic field will be opposite to the applied magnetic field. The applied magnetic field is into the plane of figure and thus the induced magnetic field is out of the plane of figure. Due to that reason, the current would be counterclockwise. So the statement is false.

C) When the loop is leaving the field, the magnetic flux through the loop will decrease. Thus, induced magnetic field will try to increase the magnetic flux i.e. the inducued magnetic field will be in the same direction as the applied magnetic field. The applied magnetic field is into the plane of figure and thus the induced magnetic field is also into the plane of figure. Due to that reason, the current would be clockwise. So the statement is true.

D) False because when entering the field magnetic force will be toward left side

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Answer:

32 cm

Explanation:

f = focal length of the converging lens = 16 cm

Since the lens produce the image with same size as object, magnification is given as

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magnification is given as

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q = p                                    eq-1

Using the lens equation, we get

1/p + 1/q = 1/f

using eq-1

1/p + 1/p = 1/16

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Answer:

\frac{R}{1} = \frac{44}{9}\ohm

Explanation:

Let us imagine that there are three wire of length equal length having equal resistances each of 44/3 Ω

Now connect these wires in parallel to so that their equivalent resistance is R.

then

\frac{1}{R} = \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}

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Correct matching:


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Answer:

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Explanation:

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