Two coils have the same number of circular turns and carry the same current. Each rotates in a magnetic field acting perpendicularly to its axis of rotation. Coil 1 has a radius of 4.5 cm and rotates in a 0.21-T field. Coil 2 rotates in a 0.39-T field. Each coil experiences the same maximum torque. What is the radius (in cm) of coil 2

Answers

Answer 1

Answer:

Explanation:

Torque acting on a coil in a magnetic field = MBsinθ where M is magnetic moment , B is magnetic field and θ is inclination of the normal to coil with direction of field.

For maximum torque sinθ = 1

Maximum torque = MB

M = NIA where N is no of turns , I is current and A is area of the coil

Maximum torque = NIAB

As maximum torque is same

N₁I₁A₁B₁ = N₂I₂A₂B₂

N₁ = N₂ , I₁ = I₂

A₁B₁ = A₂B₂

π R₁² B₁ = π R₂² B₂

4.5² x .21 = R₂² x .39

R₂² = 10.9

R₂ = 3.3 cm .


Related Questions

The skater lowers her arms as shown in the adjacent
figure decreasing her radius to 0.15 m. Find her new speed.

Answers

Answer:

is there more?

Explanation:

On a sunny day, a rooftop solar panel delivers 60 W of power to the house at an emf of 17 V. How much current flows through the panel

Answers

Answer:

3.53 amps

Explanation:

Given data

Power= 60W

Voltage= 17V

The expression relating current, power, and voltage is

P= IV

substitute

60= I*17

I= 60/17

I= 3.53 amps

Hence the current that flows is 3.53 amps

The average marathon runner can complete the 42.2-km distance of the marathon in 3 h and 30 min. If the runner's mass is 85 kg, what is the runner's average kinetic energy during the run

Answers

Answer:

the runner's average kinetic energy during the run is 476.96 J.

Explanation:

Given;

mass of the runner, m = 85 kg

distance covered by the runner, d = 42.2 km = 42,200 m

time to complete the race, t = 3 hours 30 mins = (3 x 3600s) + (30 x 60s)

                                                                               = 12,600 s

The speed of the runner, v = d/t

                                          v = 42,200 / 12,600

                                          v = 3.35 m/s

The runner's average kinetic energy during the run is calculated as;

K.E = ¹/₂mv²

K.E = ¹/₂ × 85 × (3.35)²

K.E = 476.96 J

Therefore, the runner's average kinetic energy during the run is 476.96 J.

which form of energy is an example of kinetic energy

Answers

Answer:

1. realizing of arrow

2. kicking of ball

3. punching the punching bag

The water side of the wall of a 60-m-long dam is a quarter-circle with a radius of 7 m. Determine the hydrostatic force on the dam and its line of action when the dam is filled to the rim. Take the density of water to be 1000 kg/m3.

Answers

Answer:

[tex]26852726.19\ \text{N}[/tex]

[tex]57.52^{\circ}[/tex]

Explanation:

r = Radius of circle = 7 m

w = Width of dam = 60 m

h = Height of the dam will be half the radius = [tex]\dfrac{r}{2}[/tex]

A = Area = [tex]rw[/tex]

V = Volume = [tex]w\dfrac{\pi r^2}{4}[/tex]

Horizontal force is given by

[tex]F_x=\rho ghA\\\Rightarrow F_x=1000\times 9.81\times \dfrac{7}{2}\times  7\times 60\\\Rightarrow F_x=14420700\ \text{N}[/tex]

Vertical force is given by

[tex]F_y=\rho gV\\\Rightarrow F_y=1000\times 9.81\times 60\times \dfrac{\pi 7^2}{4}\\\Rightarrow F_y=22651982.59\ \text{N}[/tex]

Resultant force is

[tex]F=\sqrt{F_x^2+F_y^2}\\\Rightarrow F=\sqrt{14420700^2+22651982.59^2}\\\Rightarrow F=26852726.19\ \text{N}[/tex]

The hydrostatic force on the dam is [tex]26852726.19\ \text{N}[/tex].

The direction is given by

[tex]\theta=\tan^{-1}\dfrac{F_y}{F_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{22651982.59}{14420700}\\\Rightarrow \theta=57.52^{\circ}[/tex]

The line of action is [tex]57.52^{\circ}[/tex].

student builds a simple circuit with a single resistor with resistance R and measures an electric potential difference ΔV across the resistor. Then, the student replaces the resistor with a new one of resistance 2R and keeps the electric potential difference the same. What is the current (I2) through the new circuit in terms of the original current​(I1)?              6mks

Answers

Answer:

 i₂ = ½ i₁

Explanation:

Let's write ohm's law for the first statement

           V = i₁ R

            i₁ = V / R

tells us that we change the resistance to R '= 2R

let's write ohm's law

            V = i₂ R’

indicates that V remains constant

            i₂ = V / R '

            i₂ = V / 2R

            i₂ = ½ V/R

we substitute

            i₂ = ½ i₁

Potential energy is





stored in a state of readiness.


conserved.


energy of motion.


not measurable.

Answers

Answer:

stored in a state of readiness.

Explanation:

Potential energy is stored in a state of readiness.

Potential energy can be defined as an energy possessed by an object or body due to its position.

Mathematically, potential energy is given by the formula;

[tex] P.E = mgh[/tex]

Where, P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

How does the size of a wind turbine affect its energy output?
A.)Smaller turbines spin slower.
B.) Larger turbines have a greater storage capacity.
C.) Larger turbines generate more electricity.
D.)Smaller turbines are better for capturing strong winds.

Answers

Answer:

Larger tubines generate more electricity.

Explanation:

Larger blades allow the turbine to capture more of the kinetic energy of the wind by moving more air through the rotors. However, larger blades require more space and higher wind speeds to operate. This distance is necessary to avoid interference between turbines, which decreases the power output.

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