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4 years ago

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A copper wire 12 cm long and a mass of 16 g is in a magnetic field of 1.2 T oriented perpendicular to the wire. What current wou

ld create a force strong enough to levitate the wire?

1 answer:

Ugo [173]4 years ago

8 0

Answer:

Current, I = 1.08 A

Explanation:

Given that,

Length of the copper wire, l = 12 cm = 0.12 m

Mass of the wire, m = 16 g = 0.016 kg

Magnetic field, B = 1.2 T

The magnetic field is oriented perpendicular to the wire. We need to find the current in the wire.

Magnetic force, F = BIL

Force of gravity, F' = mg

F' = F

$BIL=mg$

$I=\dfrac{mg}{BL}$

$I=\dfrac{0.016\times 9.8}{1.2\times 0.12}$

I = 1.08 A

So, the current through the wire is 1.08 A. Hence, this is the required solution.

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levacccp [35]

Answer:

3.82746e+26 watts

Explanation:

There are two ways to solve this problem. One way is to use the equation

L = 4πσR²T⁴

where

L = the sun's bolometric (all-spectrum) luminous power

σ = 5.670374419e-8 W m⁻² K⁻⁴ = the Stefan-Boltzmann constant

R = 6.957e+8 meters = the sun's radius

T = 5771.8 K = the sun's effective temperature

You find that

L = 3.82746e+26 watts

The other way to solve the problem is to use the Planck integral for radiant flux.

L = 4π²R ∫(v₁,v₂) 2hv³/{c² exp[hv/(kT)]−1} dv

where

h = 6.62607015e-34 J sec

c = 299792458 m sec⁻¹

k = 1.380649e-23 J K⁻¹

v₁ = 0 = frequency band lower bound, in Hz

v₂ = ∞ = frequency band upper bound, in Hz

You find, once again, that

L = 3.82746e+26 watts

The advantage of using the Planck integral becomes clear when you want to calculate the sun's luminous power only in a specific band, rather than across the entire spectrum. For example, if we do the calculation again, except that we use

v₁ = 4.1e+14 = frequency band lower bound, in Hz

v₂ = 7.7e+14 Hz = frequency band upper bound, in Hz

restricting ourselves to the visible spectrum. We find that

L (visible) = 1.56799e+26 watts

So the fraction of the sun's luminosity that is in the visible spectrum is

L (visible) / L = 0.4096686

5 0

4 years ago

Dolphins rely on echolocation to be able to survive in the ocean. In a 20 °C ocean, a dolphin produces an ultrasonic sound with

Firdavs [7]

Answer:

wavelength = 0.01 m

distance = 162.8 m

Explanation:

Given that;

Speed of sound in water = 1,480 meters per second

Frequency of ultrasound = 125KHZ

From=

v=λf

v= speed of sound

λ= wavelength of sound

f= frequency of sound

λ= 1,480 ms-1/125 * 10^3 Hz

λ= 0.01 m

From

v = 2x/t

where;

v= velocity of sound in water

x= distance traveled

t = time taken

x = vt/2

x = 1,480 ms-1 * 0.220 s/2

x= 162.8 m

8 0

3 years ago

Point A of the circular disk is at the angular position θ = 0 at time t = 0. The disk has angular velocity ω0 = 0.17 rad/s at t

statuscvo [17]

Given that,

Angular velocity = 0.17 rad/s

Angular acceleration = 1.3 rad/s²

Time = 1.7 s

We need to calculate the angular velocity

Using angular equation of motion

$\omega=\omega_{0}+\alpha t$

Put the value in the equation

$\omega=0.17+1.3\times1.7$

$\omega=2.38(k)\ m/s$

We need to calculate the angular displacement

Using angular equation of motion

$\theta=\theta_{0}+\omega_{0}t+\dfrac{\alpha t^2}{2}$

Put the value in the equation

$\theta=0+0.17\times1.7+\dfrac{1.3\times1.7^2}{2}$

$\theta=2.1675\times\dfrac{180}{\pi}$

$\theta= 124.18^{\circ}$

We need to calculate the velocity at point A

Using equation of motion

$v_{A}=v_{0}+\omega\times r$

Put the value into the formula

$v_{A}=0+2.38(k) \times0.2(\cos(124.18)i+\sin(124.18)j))$

$v_{A}=0.476\cos(124.18)j+0.476\sin(124.18)i$

$v_{A}=(-0.267j-0.393i)\ m/s$

We need to calculate the acceleration at point A

Using equation of motion

$a_{A}=a_{0}+\alpha\times r+\omega\times(\omega\times r)$

Put the value in the equation

$a_{A}=0+1.3(k)\times0.2(\cos(124.18)i+\sin(124.18)j)+2.38\times2.38\times0.2(\cos(124.18)i+\sin(124.18)j)$

$a_{A}=0.26\cos(124.18)i+0.26\sin(124.18)j+(2.38)^2\times0.2(\cos(124.18)i+\sin(124.18)j)$

$a_{A}=-0.146j-0.215i−0.636i+0.937j$

$a_{A}=0.791j-0.851i$

$a_{A}=-0.851i+0.791j\ m/s^2$

Hence, (a). The velocity at point A is $(-0.267j-0.393i)\ m/s$

(b). The acceleration at point A is $(-0.851i+0.791j)\ m/s^2$

3 0

4 years ago

A car traveled at an average speed of 60 mph for two hours. How far did it travel? 30 miles

bulgar [2K]

Answer:

120 miles

Explanation:

Speed = Δd/Δt

60 = d/2

Distance = 120 miles

8 0

3 years ago

Read 2 more answers

Thirty five soccer players were challenged to do as many sit-ups possible in two minutes

Ghella [55]

Answer:

Explanation:

What is the question?

7 0

3 years ago