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zlopas [31]
3 years ago
5

A rectangular loop with dimensions 4.20 cm by 9.50 cm carries current I. The current in the loop produces a magnetic field at th

e center of the loop that has magnitude 3.10×10−5 T and direction away from you as you view the plane of the loop. Part A What is the magnitude of the current in the loop?
Physics
1 answer:
tiny-mole [99]3 years ago
8 0

Answer:

I=1.48 A

Explanation:

Given that

B=3.1 x 10⁻5 T

b= 4.2 cm

l= 9.5 cm

The relationship for magnetic field  and current given as

B=\dfrac{2\mu _oI}{\pi}D

Where

D=\dfrac{\sqrt{l^2+b^2}}{lb}

By putting the values

D=\dfrac{\sqrt{l^2+b^2}}{lb}

D=\dfrac{\sqrt{0.042^2+0.095^2}}{0.042\times 0.095}

D=26.03 m⁻¹

B=\dfrac{2\mu _oI}{\pi}D

3.1\times 10^{-5}=\dfrac{2\times 4\times \pi \times 10^{-7} I}{\pi}\times 26.03

I=\dfrac{3.1\times 10^{-5}}{{2\times 4\times 10^{-7} }\times 26.03}

I=1.48 A

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ValentinkaMS [17]

Answer:

the value of the final pressure is 0.168 atm

Explanation:

Given the data in the question;

Let p₁ be initial pressure, v₁ be initial volume.

After expansion, p₂ is final pressure and v₂ is final volume.

So using the following equations;

p₁v₁ = nRT

p₂v₂ = nRT

hence, p₁v₁ = p₂v₂

we find p₂

p₂ = p₁v₁ / v₂

given that; initial volume v₁ = 0.175 m³, Initial pressure p₁ = 0.350 atm,

final volume v₂ = 0.365 m³

we substitute

p₂ = ( 0.350 atm × 0.175 m³ ) / 0.365 m³

p₂ = 0.06125 atm-m³ / 0.365 m³

p₂ = 0.168 atm

Therefore, the value of the final pressure is 0.168 atm

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3 years ago
(06.04 MC)<br> Which of the following best describes a human egg cell? (3 points)
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Transform boundaries are classified under which type of fault?
cluponka [151]

Answer:

Strike-slip fault

Explanation:

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When the plates are rubbed against each other, they result in enormous amount of stresses which leads to the breaking of the part of a rock causing earthquakes. Places of occurrence of these breaks are termed as faults.

Strike slip faults results from compression which takes place horizontally, but but in this the rock displacement  releases energy and takes place in a horizontal direction which is parallel to the force of compression.

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Please answer quick for brainlist ; )
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The diagram assigned B

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An alpha particle has a charge of +2e and a mass of 6.64 x 10-27 kg. It is accelerated from rest through a potential difference
kondor19780726 [428]

Answer:

a) v = 1.075*10^7 m/s

b) FB = 7.57*10^-12 N

c) r = 10.1 cm

Explanation:

(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:

K=qV          (1)

q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C

V: potential difference = 1.2*10^6 V

You replace the values of the parameters in the equation (1):

K=(3.2*10^{-19}C)(1.2*10^6V)=3.84*10^{-13}J

The kinetic energy of the particle is also:

K=\frac{1}{2}mv^2       (2)

m: mass of the particle = 6.64*10^⁻27 kg

You solve the last equation for v:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.84*10^{-13}J)}{6.64*10^{-27}kg}}\\\\v=1.075*10^7\frac{m}{s}

the sped of the alpha particle is 1.075*10^6 m/s

b) The magnetic force on the particle is given by:

|F_B|=qvBsin(\theta)

B: magnitude of the magnetic field = 2.2 T

The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1

|F_B|=(3.2*10^{-19}C)(1.075*10^6m/s)(2.2T)=7.57*10^{-12}N

the force exerted by the magnetic field on the particle is 7.57*10^-12 N

c) The particle describes a circumference with a radius given by:

r=\frac{mv}{qB}=\frac{(6.64*10^{-27}kg)(1.075*10^7m/s)}{(3.2*10^{-19}C)(2.2T)}\\\\r=0.101m=10.1cm

the radius of the trajectory of the electron is 10.1 cm

6 0
3 years ago
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