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tankabanditka [31]
3 years ago
7

Why does a torque rod have a stronger magnetic field than a torque coil?

Physics
1 answer:
Umnica [9.8K]3 years ago
8 0

Magnetic torquer or torque rod are essentially sets of electromagnets that are laid out to create a magnetic field that interfaces with an ambient magnetic field, so that the counter-forces produced is a stronger magnetic force. In addition, its magnetic field strength increases with the current due to the rod that has field lines pointing in the same direction, closed together and nearly parallel with uniformly spaced. This set up indicates that the field of a torque rod produced powerful magnetic field than a torque coil.

 

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NeX [460]

Answer:

A

Explanation:

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The kinetic energy of an object is increased by a factor of 4 . By what factor is the magnitude of its momentum changed?(a) 16(b
maw [93]

The kinetic energy of an object is increased by a factor of 4 . By what factor is the magnitude of its momentum changed: 2.

<h3>What is kinetic energy?</h3>
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The kinetic energy of an object is increased by a factor of 4 . By what factor is the magnitude of its momentum changed: 2.

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brainly.com/question/25959744

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4 0
1 year ago
A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t
lutik1710 [3]

a) E = 0

b) 3.38\cdot 10^6 N/C

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

\int EdS=\frac{q}{\epsilon_0}

where

E is the electric field

q is the charge contained by the Gaussian surface

\epsilon_0 is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

r_1 = 10 cm\\r_2 = 15 cm

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

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b)

Here we want to find the magnitude of the electric field at a distance of

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from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

E=\frac{Q}{4\pi \epsilon_0 r^2}

where in this problem:

Q=15 \mu C = 15\cdot 10^{-6} C is the charge on the shell

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Substituting, we find:

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