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3 years ago

What is the magnetic field at the center of a circular loop

Physics

1 answer:

Elan Coil [88]3 years ago

7 0

Answer:

The magnetic field at the center of a circular loop is $3.14\times10^{-5}\ T$ .

Explanation:

Given that,

Radius = 4.0 cm

Current = 2.0 A

We need to calculate the magnetic field at the center of a circular loop

Using formula of magnetic field

$B = \dfrac{I\mu_{0}}{2r}$

Where, I = current

r = radius

Put the value into the formula

$B =\dfrac{2.0\times4\pi\times10^{-7}}{2\times4.0\times10^{-2}}$

$B =0.00003141\ T$

$B=3.14\times10^{-5}\ T$

Hence, The magnetic field at the center of a circular loop is $3.14\times10^{-5}\ T$ .

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A "590-W" electric heater is designed to operate from 120-V lines.

lukranit [14]

Answer:

a) 24.4 Ω

b) 4.92 A

c) 495.9 W

c. It will be larger. The resistance will be smaller so the current drawn will increase, increasing the power.

Explanation:

The formula for power is:

P = IV

where,

P = Power of heater = 590 W

V = Voltage it takes = 120 V

I = Current Drawn = ?

Therefore,

590 W = (I)(120 V)

I = 590 W/120 V

I = 4.92 A

From Ohm's Law:

V = IR

R = V/I

Therefore,

R = 120 V/4.92 A

R = 24.4 Ω

For constant resistance and 110 V the power becomes:

P = V²/R

Therefore,

P = (110 V)²/24.4 Ω

P = 495.9 W

If the resistance decreases, it will increase the current according to Ohm's Law. As a result of increase in current the power shall increase according to formula (P = VI). Therefore, correct option is:

c. It will be larger. The resistance will be smaller so the current drawn will increase, increasing the power.

7 0

2 years ago

A 2000 g of C 14 is left to decay radioactively the half-life of Corbin 14 is approximately 5700 years what fraction of that sam

ahrayia [7]

Answer:

1/8

Explanation:

17,100 years is 3 times the half-life of 5,700 years. After each half-life, half remains, so the amount remaining after 3 half-lives is ...

(1/2)(1/2)(1/2) = 1/8

1/8 of the sample remains after 17,100 years.

8 0

2 years ago

Read 2 more answers

Inside most ball-point pens is a small spring that compresses as the pen is pressed against the paper. If a force of 0.1 N compr

AnnZ [28]

Answer:

20 N/m

Explanation:

From the question,

The ball-point pen obays hook's law.

From hook's law,

F = ke............................ Equation 1

Where F = Force, k = spring constant, e = compression.

Make k the subject of the equation

k = F/e........................ Equation 2

Given: F = 0.1 N, e = 0.005 m.

Substitute these values into equation 2

k = 0.1/0.005

k = 20 N/m.

Hence the spring constant of the tiny spring is 20 N/m

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2 years ago

A muon has a rest mass energy of 105.7 MeV, and it decays into an electron and a massless particle. If all the lost mass is conv

sergeinik [125]

Answer:

The electron’s velocity is 0.9999 c m/s.

Explanation:

Given that,

Rest mass energy of muon = 105.7 MeV

We know the rest mass of electron = 0.511 Mev

We need to calculate the value of γ

Using formula of energy

$K_{rel}=(\gamma-1)mc^2$

$\dfrac{K_{rel}}{mc^2}=\gamma-1$

Put the value into the formula

$\gamma=\dfrac{105.7}{0.511}+1$

$\gamma=208$

We need to calculate the electron’s velocity

Using formula of velocity

$\gamma=\dfrac{1}{\sqrt{1-(\dfrac{v}{c})^2}}$

$\gamma^2=\dfrac{1}{1-\dfrac{v^2}{c^2}}$

$\gamma^2-\gamma^2\times\dfrac{v^2}{c^2}=1$

$v^2=\dfrac{1-\gamma^2}{-\gamma^2}\times c^2$

Put the value into the formula

$v^2=\dfrac{1-(208)^2}{-208^2}\times c^2$

$v=c\sqrt{\dfrac{1-(208)^2}{-208^2}}$

$v=0.9999 c\ m/s$

Hence, The electron’s velocity is 0.9999 c m/s.

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2 years ago

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juin [17]

What does this all mean sorry ignore me

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2 years ago