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A proton moving in a uniform magnetic field with v⃗ 1=1.16×106ı^m/s experiences force F⃗ 1=1.61×10−16k^N. A second proton with v⃗ 2=2.26×106ȷ^m/s experiences F⃗ 2=−3.86×10−16k^N in the same field.A. What is the magnitude of B⃗ ?B. What is the direction of B⃗ ? Give your answer as an angle measured ccw from the +x-axis

Respuesta :

Answer:

B = [(1.07 × 10⁻³î) + (8.67 × 10⁻⁴j)] T

Magnitude of B = (1.377 × 10⁻³)

Direction of B = 39° anticlockwise from the +x-axis but in the clockwise direction, the direction becomes (360° - 39°) = 321°

Explanation:

The solution to this problem is presented in the attached image to this question.

Ver imagen AyBaba7
Ver imagen AyBaba7

The expression for the magnetic force and the solution of determinants allows to find the result for the value of the magnetic field is:

  • The modulus  is:  B = 13.75 10⁻⁴ T
  • The direction  is:  θ = 39º

Given parameters

  • The speed of the proton 1 is: v₁ = 1.16 10⁶ i ^ m / s
  • The force is: F₁1 = 1.61 10⁻¹⁶ k ^
  • The velocity of the second proton is: v₂ = 2.26 10⁶ j ^
  • The force on it is: F₂ = -3.86 10⁻¹⁶ k ^

To find

  • Magnetic field

The magnetic force is given by the vector product of the velocity and the magnetic field.

         F = q v x B

Where the bold letters indicate vectors, F is the force, q the electric charge, v the speed and B the magnetic field.

The best way to find the force is by solving the determinant.

        [tex]F_x i + F_y j + F_z k = q \ \left[\begin{array}{ccc}i&j&k\\v_x&v_y&v_z\\B_x&B_y&B_z\end{array}\right][/tex]  

Let us apply this expression to our cases.

Proton 1

          [tex]0 i + 0 j + 1.61 \ 10^{-16} = 1.6 \ 10 ^{-19} \ \left[\begin{array}{ccc}i&j&k\\1.16 \ &0&0\\B_x&B_y&B_z\end{array}\right] \ 10^6[/tex]

          1.61 10⁻¹⁶ k = 1.6 10⁻¹³ ( [tex]-1.16 B_z j + 1.16 B_x k[/tex] )

let's write the components of each equation.

  •  0 = 1.16 [tex]B_z[/tex]            

          [tex]B_z[/tex] = 0

  • 1.61 10⁻¹⁶ = 1.6 10⁻¹³ 1.16 [tex]B_y[/tex]

Let's calculate .        

          1.61 10⁻¹⁶ = 1.856 10⁻¹³  [tex]B_y[/tex]y

          [tex]B_y[/tex] = [tex]\frac{1.61}{1.856} \ 10^{-3}[/tex]  

          [tex]B_y[/tex] = 8.67 10⁻⁴ T

 

Proton 2

         [tex]0 i + 0 j - 3.86 \ 10^{-16} = 1.6 \ 10^{-19} \ \left[\begin{array}{ccc}i&j&k\\0&2.26&0\\B_x&B_y&B_z\end{array}\right] \ 10^6[/tex]

                    [tex]- 3.86 10^{-16} k = 1.6 \ 10^{-13} \ (2.26 B_z i - 2.26 B_x k )[/tex]    

let's write the components of the equations.

  • 0 = 2.26 [tex]B_z[/tex]

         [tex]B_z[/tex]  = 0

  • 3.86 10⁻¹⁶ = 1.6 10⁻¹³ [tex]B_x[/tex] 2.26  

       3.86 10⁻¹⁶ = 3.616 10⁻¹³ [tex]B_x[/tex]  

       [tex]B_x = \frac{3.86}{3.616} \ 10^{-3} \\B_x = 1.067 \ 10^{-3} \ \ T[/tex]

       

Let's compose the vector of the magnetic field.

      B = (10.67 i ^ + 8.67 j ^ + 0 k ^) 10⁻⁴  T

Let's transform this result in the form of a module and an angle.

We use the Pythagorean theorem to find the modulus.

      B = [tex]\sqrt{B_x^2 + B_y^2 }[/tex]  

      B = [tex]\sqrt{10.67^2 + 8.67^2} \ \ 10^{-4}[/tex]  

      B = 13.75 10⁻⁴ T

We use trigonometry for the angle.

      Tan θ = [tex]\frac{B_y}{B_x}[/tex]  

      θ = tan⁻¹ [tex]\frac{B_y}{B_x}[/tex]  

      θ = tan⁻¹ ([tex]\frac{8.67}{10.67}[/tex]8.67 / 10.67)

      θ = 39º

In conclusion, using the expression for the magnetic force and the solution of determinants, we can find the result for the value of the magnetic field is:

  • The modulus is: B = 13.75 10⁻⁴ T
  • The direction  is: θ = 39º

Learn more here: brainly.com/question/9633443

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