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3 years ago

Generators use a magnetic field of which of the following to produce an electric current?

Physics

2 answers:

Scrat [10]3 years ago

8 0

The answer is going to be magnetic domain

Bas_tet [7]3 years ago

3 0

Answer:

The answer is: Permanent magnet.

Explanation:

The rotor can be constituted by a permanent magnet or more frequently, by an electromagnet. An electromagnet is a device formed by a coil wound around a ferromagnetic material through which a current is circulated, which produces a magnetic field.

The answer is: Permanent magnet.

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A body of mass 200gram is executing SHM with amplitude of 20mm. The magnitude of maximum force which acts upon it is 0.6N. Calcu

igor_vitrenko [27]

The value of maximum velocity will be 0.3464 m/s². Energy or work is equal to the product of force and displacement.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

Mass of the body is,(m= 200g)

Amplitude is,(A =20 mm)

Maximum force is,F= 0.6 N

To find;

Maximum velocity

Energy or work is equal to the product of force and displacement.

$\rm KE = \frac{1}{2} mV_{max}^2 \\\\ \rm F_{max} \times A = \frac{1}{2} mV_{max}^2 \\\\ F_{max}= \frac{1}{2}\frac{ mV_{max}^2}{A} \\\\ 0.6= \frac{1}{2}\times \frac{ 0.2 \times V_{max}^2}{20 \times 10^{-3}} \\\\ V_{max}=0.3464 \ m/sec^2$

Hence,the value of maximum velocity will be 0.3464 m/s².

To learn more about the velocity, refer to the link;

brainly.com/question/862972

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7 0

2 years ago

Read 2 more answers

A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-

alexdok [17]

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, $q_1=-3\ nC$

It is placed at a distance of 9 cm at x axis

Charge, $q_2=+4\ nC$

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

$\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0$

Here,

$r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}$

So,

$\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}$