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3 years ago

A coil of wire that is carrying a current and produces a magnetic field is

Physics

2 answers:

labwork [276]3 years ago

8 0

The correct choice is

a solenoid

a solenoid is a sequence of a number of loops in the form of a helix in which when current flows, a magnetic field is produced along the axis of the solenoid. The magnetic field thus produced by a solenoid is given as

B = μ₀ n i

where B = magnetic field , n = number of turns per unit length, i = current in the solenoid.

snow_tiger [21]3 years ago

4 0

Answer:

Solenoid is the correct answer

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To receive or just receiving the ball.

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3 years ago

A baseball is thrown vertically up to a height of 30 m on Earth. If the same ball is thrown up on the moon with the same initial

nekit [7.7K]

Answer:

hmax = 181.48m

Explanation:

In order to calculate the maximum height reached by the same ball in the moon, you first calculate the initial velocity of the ball by using the information about the maximum height on the Earth. You use the following formula:

$h_{max}=\frac{v_o^2}{2g}$ (1)

hmax: maximum height reached by the ball in the Earth = 30m

vo: initial velocity of the ball = ?

g: gravitational acceleration on Earth = 9.8m/s^2

From the equation (1) you solve for vo:

$v_o=\sqrt{2gh_{max}}=\sqrt{2(9.8m/s^2)(30m)}=24.24\frac{m}{s}$

Next, you use the same equation (1) but for the gravitational acceleration of the moon, which is given by:

g' = 1.62m/s^2

$h_{max}=\frac{(24.24m/s)^2}{2(1.62m/s^2)}=181.48m$

The same ball, with the same initial velocity, will reache a heigth of 181m in the moon.

6 0

3 years ago

12. A flat circular coil of wire having 200 turns and diameter 6.0 cm carries a current of 7.0 A. It is placed in a magnetic fie

IgorLugansk [536]

Answer:

The magnitude of the magnetic torque on the coil is 1.98 A.m²

Explanation:

Magnitude of magnetic torque in a flat circular coil is given as;

τ = NIASinθ

where;

N is the number of turns of the coil

I is the current in the coil

A is the area of the coil

θ is the angle of inclination of the coil and magnetic field

Given'

Number of turns, N = 200

Current, I = 7.0 A

Angle of inclination, θ = 30°

Diameter, d = 6 cm = 0.06 m

A = πd²/4 = π(0.06)²/4 = 0.002828 m²

τ = NIASinθ

τ = 200 x 7 x 0.002828 x Sin30

τ = 1.98 A.m²

Therefore, the magnitude of the magnetic torque on the coil is 1.98 A.m²

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3 years ago

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kotegsom [21]

C is your answer to the question

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2 years ago

Read 2 more answers

Suppose that the electric field in the Earth's atmosphere is E = 8.60 101 N/C, pointing downward. Determine the electric charge

asambeis [7]

Answer:

q = 3.87 x 10⁵ C

Explanation:

given,

Electric field, E = 8.60 x 10¹ = 86 N/C

radius of earth, R = 6371 Km = 6.371 x 10⁶ m

Coulomb constant, K = 9 x 10⁹ N · m²/C²

Charge on the earth = ?

the electric field at the point

$E =\dfrac{kq}{r^2}$

$q =\dfrac{Er^2}{k}$

inserting all the values

$q =\dfrac{86\times (6.371\times 10^6)^2}{9\times 10^{9}}$

q = 3.87 x 10⁵ C

The electric charge on the earth is equal to 3.87 x 10⁵ C

4 0

3 years ago