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Oksana_A [137]
3 years ago
15

Two straight wires are in parallel and carry electrical currents in opposite directions with the same magnitude of 2.0A. The dis

tance between the two wires 5.0 m m. If one of the wires is infinitely long and the other the wire is 0.3 m long, what is the magnitude of the force between the two wires?
Physics
1 answer:
Veronika [31]3 years ago
3 0

Answer:

Explanation:

Two straight wires

Have current in opposite direction

i1=i2=i=2Amps

Distance between two wires

r=5mm=0.005m

Length of one wire is ∞

Length of second wire is 0.3m

Force between the wire,

The force between two parallel currents I1 and I2, separated by a distance r, has a magnitude per unit length given by

F/l = μoi1i2/2πr

F/l=μoi²/2πr

μo=4π×10^-7 H/m

The force is attractive if the currents are in the same direction, repulsive if they are in opposite directions.

F/l = μoi1i2/2πr

F/0.3=4π×10^-7×2²/2π•0.005

F/0.3=1.6×10^-4

Cross multiply

F=1.6×10^-4×0.3

F=4.8×10^-5N

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Tcecarenko [31]
The weight of the meterstick is:
W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance 
d_1 = 0.50 m - 0.40 m=0.10 m
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
(mg) d_2 = 0.20 Nm
from which we find the value of d2:
d_2 =  \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
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3 years ago
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Answer:

Explanation:

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3 years ago
The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is sudde
boyakko [2]

Explanation:

Given data:

Area A = 10 cm×2 cm = 20×10⁻⁴ m²

Distance d between the plates = 1 mm = 1×10⁻³m

Voltage of the battery is emf = 100 V

Resistance = 1025 ohm

Solution:

In RC circuit, the voltage between the plates is related to time t. Initially the voltage is equal to that of battery V₀ = emf = 100V. But After time t the resistance and capacitor changes it and the final voltage is V that is given by

V = V_{0}(1-e^{\frac{-t}{RC} }  )\\\frac{V}{V_{0} } = 1-e(^{\frac{-t}{RC} })  \\e^{\frac{-t}{RC} } = 1- \frac{V}{V_{0} }

Taking natural log on both sides,

e^{\frac{-t}{RC} } = 1- \frac{V}{V_{0} }  \\\frac{-t}{RC} = ln(1-\frac{V}{V_{0} } )\\t = -RCln(1 - \frac{V}{V_{0} })

t = -RC ln (1-\frac{V}{V_{0} })        (1)

Now we can calculate the capacitance by using the area of the plates.

C = ε₀A/d

  = \frac{(8.85*10^{-12))} (20*10^{-4}) }{1*10^{-3} }

  = 18×10⁻¹²F

Now we can get the time when the voltage drop from 100 to 55 V by putting the values of C, V₀, V and R in the equation (1)

t = -RC ln (1-\frac{V}{V_{0} })

 = -(1025Ω)(18×10⁻¹² F) ln( 1 - 55/100)

 = 15×10⁻⁹s

= 15 ns

5 0
3 years ago
An AA battery is connected to a parallel-plate capacitor having 3.9cm×3.9cm plates spaced 1.5 mm apart. How much charge does the
djyliett [7]

Answer:

Q = 1.35*10⁻¹¹ C.

Explanation:

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C= \frac{Q}{V}

At the same time, we can show (applying Gauss' Law to the surface of one of the plates), that the capacitance of a parallel-plate capacitor (with a dielectric of air), can be written as follows:

C = ε₀*A / d

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C = 8.97*10⁻¹² F

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Q = C* V = 8.97*10⁻¹² F* 1.5 V = 1.35*10⁻¹¹ C

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2 years ago
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