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3 years ago

15

A 100 A current circulates around a 1.60-mm-diameter superconducting ring. What is the on-axis magnetic field strength 5.40 cm f

rom the ring? Express your answer with the appropriate units.

1 answer:

Aleksandr [31]3 years ago

6 0

Answer:

magnetic field is 2.55 × $10^{-7}$ T

Explanation:

given data

current = 100 A

diameter = 1.60 mm

strength = 5.40 cm

to find out

What is the on-axis magnetic field

solution

we know here magnetic field on x axis current carry circular loop so it will

B = u / 4π × 2π R² I / $( x^{2} + R^{2} )^{3/2}$ ...............1

here I is current and r is radius and x distance

so

B = 4π × $10^{-7}$ / 4π × 2π (1.60× 10^-3 / 2 )² (100) / $( (5.40*10^{-2})^{2} + (1.60*10^{-3}/ 2 )^{2} )^{3/2}$

solve it and we get B

B = 2.55 × $10^{-7}$ T

magnetic field is 2.55 × $10^{-7}$ T

You might be interested in

Write two different unit in which mass is measured.

inessss [21]

Answer:

kilograms and grams

Explanation:

kilograms is the stadard unit for mass according to the SI system.

Grams is another unit for mass.

5 0

3 years ago

A 100-watt lightbulb radiates energy at a rate of 100 J/s. (The watt, a unit of power, or energy over time, is defined as 1 J/s.

fomenos

Answer:

$N=2.18\cdot 10^{20}photons/s$

Explanation:

The energy of a photon is: $E=h\nu=h\frac{c}{\lambda}$ when h is the Planck constant.

$h=6.63\cdot 10^{-34}Js$

$c=3.8\cdot 10^{8}m/s$

Then, we calculate the energy of a photon with a wavelength equal to 550 nm.

$E= 4.58\cdot 10^{-19}J$

To find the number of photons we can use this equation:

$N=\frac{P}{E}=\frac{100}{4.58\cdot 10^{-19}}=2.18\cdot 10^{20} photons/s$

Have a nice day!

7 0

3 years ago

A 63-kg hiker is climbing the 828-m-tall Burj Khalifa in Dubai. If the efficiency of converting the energy content of the bars i

konstantin123 [22]

Answer:

$T_f=5854.76$ °C

Explanation:

Given:

mass of hiker, m= 63 kg

height to be climbed, h= 828 m

energy produced by an energy bar, $E= 1.10\times 10^6 J$

heat capacity of the hiker, $c=75.3 J.mol^{-1}.K^{-1}= 4.184 J.kg^{-1}.K^{-1}$

initial body temperature of hiker, $T_i=36.6 \degree C$

<em>The efficiency of converting the energy content of the bars into the work of climbing is 25%, the remaining 75% of the energy released through metabolism is heat released to her body.</em>

We find the energy required for climbing 828 m height:

W=m.g.h

$W=63\times 9.8\times 828$

W= 511207.2 J

∵Hike eats 2 energy bars= $2\times 1.1\times 10^{6} J$

Energy produced $= 2.2\times 10^{6} J$

Now, according to her efficiency:

Total energy required for producing the work of W= 511207.2 J which is required to climb the given height will be (say, E):

$25\% of E= 511207.2$

$\Rightarrow E= 511207.2\times \frac{100}{25}$

$E=2044828.8 J$

&

Amount of total energy (E) converted into heat(Q) is:

$Q=2044828.8-511207.2\\Q=1533621.6J$

As we know that:

heat, $Q=m.c. (T_f-T_i)$ .................(1)

where:

$T_f$ is the final temperature

Putting respective values in the eq. (1)

$1533621.6= 63\times 4.184\times (T_f-36.6)$

$(T_f-36.6)\approx 5818.16$

$T_f\approx 5854.76$ °C

4 0

3 years ago

Does the charge that flows into the capacitor during the charging go all the way through the capacitor and back to the battery,

snow_lady [41]

The capacitor is used to store electric charge.That is what makes capacitors special. <span>
The charge that flows into the capacitor is stored on the plate of the capacitor that the source voltage is connected to. </span>When current flows into a capacitor, the charges get “stuck” on the plates because they can’t get past the insulating dielectric. One plate is positively charged and the other negatively <span>The stationary charges on these plates create an </span>electric field. <span>When charges group together on a capacitor like this, the cap is storing electric energy just as a battery might store chemical energy.</span>

8 0

4 years ago

Two horizontal forces are exerted on a large crate. The first force is 317 N to the right. The second force is 173 N to the left

DENIUS [597]

Answer:

A. Refer to the figure.

B. The net force acting on the crate is

$F_{net} = F_1 - F_2 = 317 - 173 = 144N$

C. We should first find the acceleration of the crate using the following kinematics equation:

$v = v_0 + at\\6.5 = 0 + 5a\\a = 1.3 m/s^2$

Mass of the crate can be found by Newton’s Second Law:

$F = Ma\\M = F/a = 144/1.3 = 110.7 kg$

7 0

4 years ago