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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

Answer :

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.

${teks-lihat-gambar} AyBaba7
${teks-lihat-gambar} 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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