A circular loop of radius 13 cm carries a current of 13 A. A flat coil of radius 0.94 cm, having 58 turns and a current of 1.9 A, is concentric with the loop. The plane of the loop is perpendicular to the plane of the coil. Assume the loop's magnetic field is uniform across the coil. What is the magnitude of (a) the magnetic field produced by the loop at its center and (b) the torque on the coil due to the loop

Answers

Answer 1

Answer:

[tex]6.28\times 10^{-5}\ \text{T}[/tex]

[tex]1.92\times 10^{-6}\ \text{Nm}[/tex]

Explanation:

[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi 10^{-7}\ \text{H/m}[/tex]

[tex]I_l[/tex] = Current in circular loop = 13 A

[tex]r_l[/tex] = Radius of circular loop = 13 cm

[tex]N[/tex] = Number of turns = 58

[tex]r_c[/tex] = Radius of coil = 0.94 cm

[tex]I_c[/tex] = Current in coil = 1.9 A

[tex]\theta[/tex] = Angle between loop and coil = [tex]90^{\circ}[/tex]

Magnitude of magnetic field in circular loop

[tex]B_l=\dfrac{\mu_0I_l}{2r_l}\\\Rightarrow B_l=\dfrac{4\pi 10^{-7}\times 13}{2\times 13\times 10^{-2}}\\\Rightarrow B_l=6.28\times 10^{-5}\ \text{T}[/tex]

The magnetic field produced by the loop at its center is [tex]6.28\times 10^{-5}\ \text{T}[/tex].

Torque is given by

[tex]\tau=\pi NI_cr_c^2B_l\sin\theta\\\Rightarrow \tau=\pi 58\times 1.9\times (0.94\times 10^{-2})^2\times 6.28\times 10^{-5}\sin90^{\circ}\\\Rightarrow \tau=1.92\times 10^{-6}\ \text{Nm}[/tex]

The torque on the coil due to the loop [tex]1.92\times 10^{-6}\ \text{Nm}[/tex].


Related Questions

According to Newton’s second law of motion, weight depends on an object’s mass and the net force acting on the object. True or false

Answers

Answer:

true..

Explanation:

Newton's 2nd law of motion is greater the mass of an object the more force it will take to accelerate the object

According to Newton’s second law of motion, weight depends on an object’s mass and the net force acting on the object. The given statement is true.

What is force?

Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body.

The following equation help to understand the statement;

Force is defined as the product of mass and acceleration. Its unit is Newton.

F=ma

F=mg

W=mg

The weight of an item is determined by its mass and the net force exerted on it, according to Newton's second rule of motion. The provided assertion is correct.

Hence,the given statement is true.

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

https://brainly.com/question/26115859

#SPJ2

1 ) when a ball is projected upwords its time of rising is ...............the time of falling .
a) greater than b) smaller than c) equal to d ) double
2 ) when an object falls freely under the effect of gravity , the distance moved is
a ) directly proportional to time
b ) inversely proportional to time
c ) directly proportional to square of time
d ) inversely proportional to square of time.

Answers

Answer:

correct answer is C

Explanation:

In this exercise, you are asked to complete the sentences so that the sentence makes sense.

1) in projectile launching, the only force that acts is gravity in the vertical direction, so the time of going up is EQUAL to the time of going down

correct answer C

2) when a body falls freely, the acceleration is the ratio of gravity, therefore if it starts from rest, its height is

            y = v₀ t - ½ gt²

v₀ = 0

             y = -1/2 g t²

so the position is not proportional to the square of the time

correct answer is C


Which wave type could move in the empty space between the Sun and the Earth

Answers

Answer:

Electromagnetic waves


Find the velocity of an object that has a mass of 500x10-6 kg and a charge of 1pc if it starts from
rest and passes through a potential difference of 120kV

Answers

Answer:

 v = 4.8 10⁻⁴ m / s

Explanation:

To solve this exercise we can use the concepts of energy. In this case the potential energy is transformed into kinetic energy

          U = K

          q V = ½ m v²

          v = [tex]\sqrt { \frac{2qV}{m} }[/tex]

in the exercise they indicate the value of the charge q₁ = 1 pC = 1 10⁻¹² C

let's calculate

          v = [tex]\sqrt{ \frac{2 \ 1 \ 10^{-12} 120 \ 10^3}{500 ^{2} }[/tex]

          v = 4.8 10⁻⁴ m / s

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