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jeka57 [31]
3 years ago
14

A long wire carrying a 5.0 A current perpendicular to the xy-plane intersects the x-axis at x= - 2.0 cm . A second, parallel wir

e carrying a 3.5 A current intersects the x-axis at x=+ 2.0cm .
A). At what point on the x-axis is the magnetic field zero if the two currents are in the same direction?

B). At what point on the x-axis is the magnetic field zero if the two currents are in opposite directions?
Physics
1 answer:
mario62 [17]3 years ago
8 0

Answer:

a . 0.35cm

b.  11.33cm

Explanation:

a. Given both currents are in the same direction, the null point lies in between them. Let x be distance of N from first wire, then distance from 2nd wire is 4-x

#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in between the wires:

\frac{\mu_oi_1}{2\pi x}=\frac{\mu_oi_2}{2\pi(4-x)}\\\\5/x=\frac{3.5}{4-x}\\\\x=2.35cm\\\\N=2.35-2=0.35cm

Hence, for currents in same direction, the point is 0.35cm

b. Given both currents flow in opposite directions, the null point lies on the other side.

#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in outside the wires:

Let x be distance of N from first wire, then distance from 2nd wire is 4+x:

\frac{\mu_oi_1}{2\pi(4+ x)}=\frac{\mu_oi_2}{2\pi x}\\\\5/(4+x)=\frac{3.5}{x}\\\\x=9.33cm\\\\N=9.33+2=11.33cm

Hence, if currents are in opposite directions the point on x-axis is 11.33cm

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(a) 112.5 V/m

(b) 0.01494 T

Explanation:

velocity, v = 4.5 km/s = 4500 m/s

magnetic field, B = 2.5 x 10^-2 T

mass, m = 6.64 x 10^-26 kg

charge, q = 1.6 x 10^-19 C = 1.6 x 10^-19 C

(a) The force due to the magnetic field is balanced by the force due to the electric field

q E = B q v

E = B v = 2.5 x 10^-2 x 4500 = 112.5 V/m

(b) radius, r = 12.5 cm

r = \frac{mv}{B'q}

B' = \frac{mv}{qr}

B' = \frac{6.64\times10^{-26}\times4500}{1.6\times 10^{-19}\times 0.125}

B' = 0.01494 T

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What does Charles Darwin theory of natural selection state about traits
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3 years ago
You want to find out how many atoms of the isotope 65Cu are in a small sample of material. You bombard the sample with neutrons
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Answer:

a) number of copper atoms 65 (⁶⁵Cu)  is 7.692 10⁶ atoms

b) m_total Cu = 1.585 10⁹ u = 2.632 10⁻¹⁸ kg

Explanation:

a) For this exercise let's start by using the radioactive decay ratio

           N = N₀  e^{- \lambda t}o e - lambda t

The half-life time is defined as the time it takes for half of the radioactive (activated) atoms to decay, therefore after two half-lives there are

            N = ½ (½ N₀) = ¼ N₀

            N₀ = 4 N

in each decay a photon is emitted so we can use a direct rule of proportions. If an atom emits a photon it has Eo = 1,04 Mev, how many photons it has energy E = 10,000 MeV

          # _atoms = 1 atom (photon) (E / Eo)

          # _atoms = 1 10000 / 1.04

          # _atoms = 9615,4 atoms

          N₀ = 4 #_atoms

          N₀ = 4 9615,4

          N₀=  38461.6  atoms

in the exercise indicates that half of the atoms decay in this way and the other half decays directly to the base state of Zinc, so the total number of activated atoms

          N_activated = 2 # _atoms

          N_activated = 2 38461.6

          N_activated = 76923.2

also indicates that 1% = 0.01 of the nuclei is activated by neutron bombardment

          N_activated = 0.01 N_total

          N_total = N_activated / 0.01

          N_total = 76923.2 / 100

          N_total = 7.692 10⁶ atoms

so the number of copper atoms 65 (⁶⁵Cu)  is 7.692 10⁶

b) the natural abundance of copper is

  ⁶³Cu     69.17%

  ⁶⁵Cu    30.83%

Let's use a direct proportion rule. If there are 7.692 10⁶  ⁶⁵Cu that represents 30.83, how much ⁶³Cu is there that represents 69.17%

                # _63Cu = 69.17%  (7.692 10⁶    / 30.83%)

                # _63Cu = 17.258 10⁶  atom  ⁶³Cu

the total amount of comatose is

              #_total Cu = #_ 65Cu + # _63Cu

              #_total Cu = (7.692 + 17.258) 10⁶

              #_total Cu = 24.95 10⁶

the atomic mass of copper is m_Cu = 63.546 u

          m_total = #_totalCu m_Cu

          m_total = 24.95 10⁶ 63,546 u

          m_total = 1.585 10⁹ u

let's reduce to kg

           m_total Cu = 1.585 10⁹ u (1,66054 10⁻²⁷ kg / 1 u)

           m_total Cu = 2.632 10⁻¹⁸ kg

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Answer:

893 Newtons

Explanation:

m = Mass of Usain Bolt = 94 kg

a = Acceleration of Usain Bolt in the first 0.89 seconds = 9.5 m/s²

From Newton's Second law

Force

F=ma

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The average horizontal force exerted by Bolt against the ground during the first 0.890 s of the race is 893 Newtons

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