Ask question

Ask question

All categories

2 years ago

9

An electric field of 4.0 μV/m is induced at a point 2.0 cm from the axis of a long solenoid (radius = 3.0 cm, 800 turns/m). At w

hat rate is the current in the solenoid changing at this instant?

1 answer:

vagabundo [1.1K]2 years ago

6 0

Answer:

The rate of current in the solenoid is 0.398 A/s

Explanation:

Given that,

Electric field $E = 4.0\ \mu V/m$

Distance = 2.0 cm

Radius = 3.0 cm

Number of turns per unit length = 800

We need to calculate the rate of current

Using formula of electric field for solenoid

$E = \dfrac{x}{2}\mu_{0}n\dfrac{dI}{dt}$

Where, x = distance

n = number of turns per unit length

E = electric field

r = radius

Put the value into the formula

$4.0\times10^{-6}=\dfrac{2.0\times10^{-2}}{2}\times4\pi\times10^{-7}\times800\times\dfrac{dI}{dt}$

$\dfrac{dI}{dt}=\dfrac{4.0\times10^{-6}\times2}{2.0\times10^{-2}\times4\pi\times10^{-7}\times800}$

$\dfrac{dI}{dt}=0.397\ A/s$

Hence, The rate of current in the solenoid is 0.398 A/s.

You might be interested in

Trained dolphins are capable of a vertical leap of 7.0 m straight up from the surface of the water - an impressive feat. Suppose

dmitriy555 [2]

Answer:14 m

Explanation:

Given

Vertical jump make by the dolphin is given by $h=7\ m$

Suppose the dolphin jump with an initial velocity of $u$

so u is given by $u^2=2\cdot g\cdot h$

If dolphin launches at an angle $\theta$ then maximum horizontal range is given by

assuming the of Dolphin to be Projectile so range is given by

$R=\frac{u^2\sin 2\theta }{g}$

substitute the value of $u^2$

$R=\frac{2\times 9.8\times 7\sin 2\theta }{9.8}$

$R=2h\sin 2\theta$

Range will be maximum for $\theta =45^{\circ}$

thus $R_{max}=2\times 7\times 1=14\ m$

3 0

2 years ago

A particle in the first excited state of a one-dimensional infinite potential energy well (with U = 0 inside the well) has an en

nataly862011 [7]

Answer:

The energy of this particle in the ground state is E₁=1.5 eV.

Explanation:

The energy $E_{n}$ of a particle of mass <em>m</em> in the <em>n</em>th energy state of an infinite square well potential with width <em>L </em>is:

$E_{n}=\frac{n^{2}h^{2}}{8mL^{2}}$

In the ground state (n=1). In the first excited state (n=2) we are told the energy is E₂= 6.0 eV. If we replace in the above equation we get that:

$E_{1}=\frac{h^{2}}{8mL^{2}}$

$E_{2}=\frac{h^{2}}{2mL^{2}}$

So we can rewrite the energy in the ground state as:

$E_{1}=\frac{1}{4}(\frac{h^{2}}{2mL^{2}})$

$E_{1}=\frac{1}{4} E_{2}$

$E_{1}=\frac{1}{4} ( 6.0\ eV)$

Finally

$E_{1}=1.5\ eV$

6 0

2 years ago

A diver in the pike position (legs straight, hands on ankles) usually makes only one or one-and-a-half rotations. To make two or

Korolek [52]

Answer:

from the above analysis we can say that the angular velocity in the later case is more than that of the former case. This means that the number of rotation made in the truck case is more than that made in pike position.

Explanation:

This can be explained on the basis of conservation of angular momentum.

This means the initial and the final angular velocity is conserved. Consider initial position (1)in the pike and final position in the be truck position. So there inertia's will also be different.

⇒ $I_1\omega_1 = I_2\omega_2$

$\frac{I_1}{I_2} = \frac{\omega_2}{\omega_1}$

also,

$I_1= mr_1^2$

$I_2= mr_2^2$

since, $r_2^2$

$I_2^2$

therefore,

$\omega_1^2$

So, from the above analysis we can say that the angular velocity in the later case is more than that of the former case. This means that the number of rotation made in the truck case is more than that made in pike position.

6 0

2 years ago

Volcanoes are often formed at plate boundaries. This is a convergent plate boundary. From the choices listed, pick the correct d

katen-ka-za [31]

Your answer should be a

8 0

2 years ago

Read 2 more answers

A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±qseparated by a distance s. T

atroni [7]

Answer:

The magnitude of the force on the dipole due to the charge Q = $\rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.$

The magnitude of the torque on the dipole = $\rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi}\dfrac{2qQs^2}{r^3}.$

Explanation:

Given that a point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q, separated by a distance s and the charge Q is located in the plane that bisects the dipole.

The magnitude of the electric field that the dipole exerts at the position where the charge Q is held is given by

$\rm E = \dfrac{k2qs}{(r^2+s^2)^{3/2}}.$

<em>where</em>,

k is the Coulomb's constant, having value = $\dfrac{1}{4\pi \epsilon_o}$

$\epsilon_o$ is the electrical permittivity of free space.

Also, r>>s, therefore, $\rm r^2+s^2\approx r^2.$

$\rm E = \dfrac{k2qs}{(r^2)^{3/2}}=\dfrac{k2qs}{r^3}.$

The magnitude of the electric force F on a charge q placed at a point and the magnitude of the electric field E at that point are related as

$\rm F=qE$

Therefore, the electric force on the charge Q due to the dipole is given by

$\rm F=Q\dfrac{k2qs}{r^3}=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}.$

According to Newton's third law of motion, the magnitude of the force exerted by the dipole on the charge Q is same as the magnitude of the force exerted by the charge on the dipole.

Thus, the magnitude of the force on the dipole due to the charge Q = $\dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.$

The magnitude of the torque on the dipole is given by

$\rm \tau = Fs\ \sin\theta$

Since the charge Q is placed in the plane that bisects the dipole, therefore, $\theta = 90^\circ$ .

$\rm \tau = \dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}\cdot s\cdot 1=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs^2}{r^3}.$

4 0

2 years ago