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Alinara [238K]
3 years ago
7

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 left. The self inductance of the loop is negligible. True While the loop is entirely in the field, the emf in the loop is zero. False When entering the field the coil experiences a magnetic force to the left. False When leaving the field the coil experiences a magnetic force to the right. True Upon entering the field, a clockwise current flows in the loop.
Physics
1 answer:
Anvisha [2.4K]3 years ago
5 0

Answer:

False, True, False, True

Explanation:

Given a square loop of a wire having a small resistance and moving with a constant sped from the free region to a region having a region having a uniform field B.  It is then moved back to the left in the free field region.

For the following statements :

While the loop is entirely in the field, the emf in the loop is zero.  -- TRUE

The emf of the coil is zero when the coil moves into the field from a free region field to a uniform field region field.

When entering the field the coil experiences a magnetic force to the left   ---- FALSE

According to the Lenz law, a loop entering the field opposes it and moves towards the right.

When leaving the field the coil experiences a magnetic force to the right. --- TRUE

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

In the field the coil or the loop tries to oppose the magnetic field and forces the current in the loop to move in a clockwise direction.

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Two negative charges that are both -3.0 C push each other apart with a force of 19.2 N. How far apart are the two charges ?
fgiga [73]
The Coulomb's force acting between two charges is
F=k_e  \frac{q_1 q_2}{r^2}
where k_e = 8.99 \cdot 10^9 N m^{-2}C^{-2} is the Coulomb's constant, q1 and q2 the two charges, and r the distance between them.

Using q_1 = q_2 = -3 C, we can find the distance between the two charges when the force is F=19.2 N:
r=  \sqrt{k_e  \frac{q_1 q_2}{F} }=  \sqrt{ 8.99\cdot 10^9 \frac{(-3.0C)^2}{19.2 N} }=6.5 \cdot 10^4 m = 65 km
5 0
3 years ago
An electron is released from rest at a distance of 0.570 m from a large insulating sheet of charge that has uniform surface char
Lelu [443]

Explanation:

Formula to calculate the electric field of the sheet is as follows.

          E = \frac{\sigma}{2 \epsilon_{o}}

And, expression for magnitude of force exerted on the electron is as follows.

            F = Eq

So, work done by the force on electron is as follows.

           W = Fs

where,     s = distance of electron from its initial position

                  = (0.570 - 0.06) m

                  = 0.51 m

First, we will calculate the electric field as follows.

              E = \frac{\sigma}{2 \epsilon_{o}}

                 = \frac{4.60 \times 10^{-12}C/m^{2}}{2 \times 8.854 \times 10^{-12}C^{2}/N m^{2}}

                 = 0.259 N/C

Now, force will be calculated as follows.

                 F = Eq

                    = 0.259 N/C \times 1.6 \times 10^{-19} C

                    = 0.415 \times 10^{-19} N

Now, work done will be as follows.

                    W = Fs

                        = 0.415 \times 10^{-19} N \times 0.51 m

                        = 2.12 \times 10^{-20} J

Thus, we can conclude that work done on the electron by the electric field of the sheet is 2.12 \times 10^{-20} J.

3 0
4 years ago
Read 2 more answers
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