How do I solve this problem

A solenoid of radius 2.0 mm contains 100 turns of wire uniformly distributed over a length of 5.0 cm. It is located in air and c

tankabanditka [31]

Answer:

The magnetic field strength inside the solenoid is $5.026\times10^{-3}\ T$ .

Explanation:

Given that,

Radius = 2.0 mm

Length = 5.0 cm

Current = 2.0 A

Number of turns = 100

(a). We need to calculate the magnetic field strength inside the solenoid

Using formula of the magnetic field strength

Using Ampere's Law

$B=\dfrac{\mu_{0}NI}{l}$

Where, N = Number of turns

I = current

l = length

Put the value into the formula

$B=\dfrac{4\pi\times10^{-7}\times100\times2.0}{5.0\times10^{-2}}$

$B=0.005026=5.026\times10^{-3}\ T$

(b). We draw the diagram

Hence, The magnetic field strength inside the solenoid is $5.026\times10^{-3}\ T$ .

4 0

3 years ago

During a certain comet’s orbit around the Sun, its closest distance to the Sun is 0.6 AU, and its farthest distance from the Sun

Ray Of Light [21]

Answer:

At the closest point

Explanation:

We can simply answer this question by applying Kepler's 2nd law of planetary motion.

It states that:

"A line connecting the center of the Sun to any other object orbiting around it (e.g. a comet) sweeps out equal areas in equal time intervals"

In this problem, we have a comet orbiting around the Sun:

- Its closest distance from the Sun is 0.6 AU

- Its farthest distance from the Sun is 35 AU

In order for Kepler's 2nd law to be valid, the line connecting the center of the Sun to the comet must move slower when the comet is farther away (because the area swept out is proportional to the product of the distance and of the velocity: $A\propto vr$ , therefore if r is larger, then v (velocity) must be lower).

On the other hand, when the the comet is closer to the Sun the line must move faster ( $A\propto vr$ , if r is smaller, v must be higher). Therefore, the comet's orbital velocity will be the largest at the closest distance to the Sun, 0.6 A.

7 0

3 years ago

A 2.5g copper penny is given a charge of -4.0*10^-9c. how mny excess electrons are on the penny?

Lyrx [107]

Answer : The excess of electrons on the penny are, $2.5\times 10^{10}$ electrons

Solution : Given,

Total charge = $-4.0\times 10^{-9}C$

Charge on electron = $-1.6\times 10^{-19}C$

Formula used :

$\text{Excess of electrons}=\frac{\text{Total charge}}{\text{Charge of electron}}$

Now put all the given values in this formula, we get the excess of electrons present on the penny.

$\text{Excess of electrons}=\frac{\text{Total charge}}{\text{Charge of electron}}=\frac{-4.0\times 10^{-9}C}{-1.6\times 10^{-19}C/e^-}=2.5\times 10^{10}e^-$

Therefore, the excess of electrons on the penny are, $2.5\times 10^{10}$ electrons

7 0

3 years ago

Suppose you "dig a hole" 52 feet deep and place a baseball machine in the hole, so that the balls fire out at 96 ft /s and land

jok3333 [9.3K]

Answer:

76.73 ft/s

Explanation:

Let the final velocity is v.

initial velocity, u = 96 ft/s

g = 32 ft/s²

height, h = 52 feet

use third equation of motion

v² = u² - 2 gh

v² = 96 x 96 - 2 x 32 x 52

v = 76.73 ft/s

Thus, the speed of the ball as it reaches the ground is 76.73 ft/s.

3 0

3 years ago

In a double-slit experiment, the third-order maximum for light of wavelength 510 nm is located 17 mm from the central bright spo

Marysya12 [62]

Answer:

14.9 mm

Explanation:

We know dsinθ = mλ where d = separating of slit, m = order of maximum = 3 and λ = wavelength = 510nm = 510 × 10⁻⁹ m

Also tanθ = L/D where L = distance of m order fringe from central bright spot = 17 mm = 0.017 m and D = distance of screen from slit = 1.6 m

So, sinθ = mλ/d

Since θ is small, sinθ ≅ tanθ

So,

mλ/d = L/D

d = mλD/L

Substituting the values of the variables into the equation, we have

d = 3 × 510 × 10⁻⁹ m × 1.6 m/0.017 m

d = 2448 × 10⁻⁹ m²/0.017 m

d = 144000 × 10⁻⁹ m

d = 1.44 × 10⁻⁴ m

d = 0.144 × 10⁻³ m

d = 0.144 mm

Now, for the second-order maximum, m' of the 670 nm wavelength of light,

m'λ'/d = L'/D where m' = order of maximum = 2, λ' = wavelength of light = 670 nm = 670 × 10⁻⁹ m, d = slit separation = 0.144 mm = 0.144 × 10⁻³ m, L' = distance of second order maximum from central bright spot and D = distance of screen from slit = 1.6m

So, L' = m'λ'D/d

So, substituting the values of the variables into the equation, we have

L' = 2 × 670 × 10⁻⁹ m × 1.6 m/0.144 × 10⁻³ m

L' = 2144 × 10⁻⁹ m²/0.144 × 10⁻³ m

L' = 14888.89 × 10⁻⁶ m

L' = 0.01488 m

L' ≅ 0.0149 m

L' = 14.9 mm

4 0

3 years ago