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2 years ago

Examine the scenario.

2 answers:

8 0

The answer is "A magnetic field influences the motion of charged particles. The negatively-charged particles will experience a force acting against their motion, so the negatively-charged particles would slow down and stop moving."

This is because for example;

Have you ever seen a magnet? If so, you would experience the same scienario as above.

aniked [119]2 years ago

5 0

Answer:

option (c)

Explanation:

According to the Fleming's left hand rule, if we spread out thumb, fore finger and middle finger of our left hand mutually perpendicular to each other, then thumb represents the direction of force, fore finger represents the direction of magnetic field and the middle finger represents the direction of current.

According to the question, positive charge particle moves in left wards direction, i.e., current is flowing towards left, the force is acting upwards, so the direction of magnetic field is in perpendicular outwards to the plane of paper.

When a negatively charged particle moves towards left , then the direction of force is downwards.

Thus, the magnetic field influences the motion of charged particles. The negatively-charged particles will experience a force downward because positive and negative particles would experience forces in opposite directions.

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3 years ago

A long, hollow, cylindrical conductor (inner radius 3.4 mm, outer radius 7.3 mm) carries a current of 36 A distributed uniformly

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Answer:

a. $B= 9.45 \times10^{-3} T$

b. $B= 0.820 T$

c. $B= 0.0584 T$

Explanation:

First, look at the picture to understand the problem before to solve it.

a. d1 = 1.1 mm

Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:

To solve the equations we have to convert all units to those of the international system. (mm→m)

$B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\$

μ0 is the constant of proportionality

μ0=4πX10^-7 N*s2/c^2

b. d2=3.6 mm

Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:

J: current density

c: outer radius

b: inner radius

The cilinder's current is negative, as it goes on opposite direction than the wire's current.

$J= \frac {-I_{c}}{\pi(c^{2}-b^{2} ) }}$

$J=\frac{-36}{\pi(5.33\times10^{-5}-1.16\times10^{-5}) } =-274.80\times10^{3} A/m^{2}$

$B=\frac{u_{0}(I_{w}-JA_{s})}{2\pi d_{2} } \\A_{s}=\pi (d_{2}^{2}-b^2)=4.40\times10^{-6} m^2\\$

$B=\frac{6.68\times10^{-5}}{8.14\times10^{-5}} =0.820 T$

c. d3=7.4 mm

Here, the point is located out of the cilinder. Therefore, we have to consider both, the conductor's current and the wire's current as follows:

$B=\frac{u_{0}(I_w-I_c)}{2\pi d_3 } =\frac{2.011\times10^-5}{3.441\times10^{-4}} =0.0584 T$

As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.

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