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Firlakuza [10]
3 years ago
11

A point charge is moving with speed 2 × 107 m/s parallel to the x axis along strait line at y=2 m. At t = 0, the charge is at x

= 0 m. The magnitude of the magnetic field at x = 4 m is B0 at the origin. The magnitude of the magnetic field at at origin when t = 0.15 μs is?
Physics
1 answer:
Oduvanchick [21]3 years ago
4 0

Answer:

B_2  = 4B_1

Explanation:

initial distance of point from the electron is r_1 = 4 m

distance moved by electron is x

x = vt

x = 2*10^{7} *0.1*10^{-6}

x = 2 m

new distance of point from electron is r_2 = 2m

field due to moving charge is

B = \frac{ \mu_o 4vr}{4 \pi r^3}

CHARGE IS INVERSELY PROPOTIONAL TO DISTANCE FROM ELECTRON.

therefore\frac{B_1}{B_2} = [\frac{r_2}{r_1}]^2

                         = [\frac{2}{4}]^2

B_2  = 4B_1

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8 0
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In the diagram, q1= +8.0 C, q2= +3.5 C, and q3 = -2.5 C. q1 to q2 is 0.10 m, q2 to q3 is 0.15 m. What is the net force on q2? La
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Answer:

f(t) =  28,7 [N]

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The net force on +q₂  is the sum of the force of +q₁  on +q₂ ( is a repulsion force since charges of equal sign repel each other ) and the force of -q₃ on +q₂ ( is an attraction force, opposite sign charges attract each other)

The two forces have the same direction to the right of charge q₂, we have to add them

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f(t) = f₁₂ + f₃₂

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q₂ =  + 3,5 μC  then  q₂ = 3,5 *10⁻⁶ C

K = 9*10⁹  [ N*m² /C²]

f₁₂ = 9*10⁹ * 8*3,5*10⁻¹²/ 1*10⁻²   [ N*m² /C²]* C*C/m²

f₁₂ = 252*10⁻¹ [N]

f₁₂ = 25,2 [N]

f₃₂ =  9*10⁹*3,5*10⁻⁶*2,5*10⁻⁶ /(0,15)²

f₃₂ =  78,75*10⁻³/ 2,25*10⁻²

f₃₂ =  35 *10⁻¹

f₃₂ =  3,5 [N]

f(t) =  28,7 [N]

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