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Fantom [35]
3 years ago
13

A flat, circular loop has 18 turns. The radius of the loop is 15.0 cm and the current through the wire is 0.51 A. Determine the

magnitude of the magnetic field at the center of the loop (in T).
Physics
1 answer:
Ostrovityanka [42]3 years ago
6 0

Answer:

The magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻⁵ T.

Explanation:

Given;

number of turns of the flat circular loop, N = 18 turns

radius of the loop, R = 15.0 cm = 0.15 m

current through the wire, I = 0.51 A

The magnetic field through the center of the loop is given by;

B = \frac{N\mu_o I}{2R}

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

B = \frac{N\mu_o I}{2R} \\\\B = \frac{18*4\pi*10^{-7} *0.51}{2*0.15} \\\\B = 3.846 *10^{-5} \ T

Therefore, the magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻⁵ T.

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Mandarinka [93]

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A stone is dropped from the edge of a roof, and hits the ground with a velocity of -180 feet per second. How high (in feet) is t
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d = 506.25 ft

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As we know by kinematics that

v_f^2 - v_i^2 = 2 a d

here we know that initially the stone is dropped from rest from the edge of the roof

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now we have

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A fireboat is to fight fires at coastal areas by drawing seawater with a density of 1030 kg/m3 through a 10-cm-diameter pipe at
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Answer:

50.93 m/s

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Explanation:

From the question, the nozzle exit diameter = 5 cm, Radius= diameter/2= 5cm/2= 2.5cm. we can convert it to metre for unit consistency= (2.5×0.01)=

0.025m

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A= πr^2

A= π ×0.025^2

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From the question, the water is moving through the pipe at a rate of 0.1 m /s , then for the water to move through it at a seconds, it must move at

(0.1 / 1.9635 ×10^- ³ m²)

= 50.93 m/s

During the Operation of the pump, the Dynamic energy of the water= potential energy provided there is no loss during the Operation

mgh = 1/2mv²

We can make "h" subject of the formula, which is the height of required head of water

h = (1/2mv²)/mg

h= v² / 2g

h = 50.93² / (2 ×9.81)

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From the question;

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the given position of nozzle = 3 m

the total head the pump needed=(The total irreversible head loss of the system + the position of the nozzle + required head of water )

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Therefore, the required shaft power input to the pump and the water discharge velocity is 199.5 kW

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