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3 months ago

4. A 505-turn circular-loop coil with a diameter of 15.5 cm is initially aligned so that

Physics

2 answers:

Ostrovityanka [3.2K]3 months ago

8 0

The intensity of the magnetic field is $4.8\cdot 10^{-5} T$

Explanation:

According to Faraday's Law, the magnitude of the induced electromotive force (emf) in the coil corresponds to the rate of flux change that links through the coil:

$\epsilon = \frac{N\Delta \Phi}{\Delta t}$ (1)

where

N = 505 represents the number of turns in the coil

$\Delta \Phi$ is the variation in magnetic flux through the coil

$\Delta t = 2.77 ms = 2.77\cdot 10^{-3} s$ indicates the time interval

$\epsilon = 0.166 V$

Rotating the coil from perpendicular to parallel alignment with the Earth's magnetic field results in the final flux equaling zero, making the magnitude of flux change simply the initial flux:

$\Delta \Phi = B A cos \theta$

where

B denotes the intensity of the magnetic field

A signifies the area of the coil

$\theta=0^{\circ}$ represents the angle between the normal to the coil and the field

The area of the coil can be expressed as

$A=\pi r^2$

where

$r=\frac{15.5 cm}{2}=7.75 cm = 7.75\cdot 10^{-2} m$ outlines its radius

By substituting all values into equation (1) and solving for B, we obtain:

$\epsilon= \frac{NB\pi r^2 cos \theta}{\Delta t}\\B=\frac{\epsilon \Delta t}{\pi r^2 cos \theta}=\frac{(0.166)(2.77\cdot 10^{-3})}{(505)\pi (7.75\cdot 10^{-2})^2(cos 0^{\circ})}=4.8\cdot 10^{-5} T$

For further reading on magnetic fields:

Softa [3K]3 months ago

7 0

Answer:

Earth's magnetic field strength is measured at 4.825 x 10⁻⁵ T

Explanation:

Provided details;

Turns count; N = 505 turns

Diameter of the circular loop, d = 15.5 cm

Average emf, V = 0.166 V

Change over time, t = 2.77 ms

$V_{avg} =- N\frac{d \phi}{dt} -------equation (i)\\\\\phi = BACos \theta \\\\d \phi = BACos \theta_f - BACos \theta_i\\\\V_{avg.} = -N(\frac{BACos \theta_f - BACos \theta_i}{dt})\\\\V_{avg.} = N(\frac{BACos \theta_i - BACos \theta_f}{dt}) ---------equation(ii)$

Initially, the circular loop's plane is perpendicular to the Earth's magnetic field, $\theta _i = 0^o$

Finally, after the coil is rotated 90.0°, $\theta_f = 90^o$

$V_{avg.} = NBA(\frac{Cos 0 -Cos 90}{t} )\\\\V_{avg.} = \frac{NBA}{t} \\\\B = \frac{V_{avg.}*t}{NA} -------------equation(iii)\\\\But, A = \frac{\pi d^2}{4} \\\\B = \frac{4*V_{avg.}*t}{N*\pi d^2}\\\\substitute \ the \ given \ values\\\\B = \frac{4*(0.166)*(2.77*10^{-3})}{(505)*\pi* (0.155)^2} = 4.825*10^{-5} \ T$

Therefore, the calculated value of Earth's magnetic field is 4.825 x 10⁻⁵ T

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inna [3103]

Answer:

the time it takes after impact for the puck is 2.18 seconds

Explanation:

initially given information

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

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temperature of air = 15°C

to determine

time needed for the puck to reduce its speed by 10%

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we note that velocity changes from 0 to v

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implying a change in velocity is du = v

and clearance dy = h

shear stress acting on the surface is expressed as

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therefore

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substituting the values

= 1.75 × $10^{-5}$ × $\frac{v}{10^{-4}}$

= 0.175 v

the area between the air and puck is given by

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area = $\frac{\pi }{4} 0.1^{2}$

area = 7.85 × $\frac{v}{10^{-3}}$ m²

thus, the force on the puck can be represented as

Force = × area

force = 0.175 v × 7.85 × $10^{-3}$

force = 1.374 × $10^{-3}$ v

now applying Newton's second law

force = mass × acceleration

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- 1.374 × $10^{-3}$ v = $0.03 \frac{0.9v - v }{t}$

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the time needed after impact for the puck is 2.18 seconds

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3 months ago

If a 50.0-kg mass weighs 554 n on the planet saturn, calculate saturn’s radius

ValentinkaMS [3465]

Answer:

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Explanation:

Mass; m = 50 kg

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From the formula:

W = mg

This simplifies to; 554 = 50g

g = 554/50

g = 11.08 m/s²

Also, using the formula;

mg = GMm/r²

hence; g = GM/r²

Rearranging gives;

r = √(GM/g)

With G as a known constant of 6.67 × 10^(-11) Nm²/kg²

r = √(6.67 × 10^(-11) × 50/11.08)

r = 17.35 × 10^(-6) m

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3 months ago

A supersonic nozzle is also a convergent–divergent duct, which is fed by a large reservoir at the inlet to the nozzle. In the re

Softa [3030]

Answer:

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$T_2$ = Exit temperature

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