Respuesta :
Answer:
1) B = 1,358 T , 2) θ = -55.35°, 3) F = 8.04 10⁻¹⁶ N , 4) θ = 55.34°
Explanation:
The magnetic force is given by
F = q v x B
Bold indicates vectors, one way to use this equation is to look for its magnitude
F = q v B sin θ
The direction is given by the right hand for a positive charge where the thumb points in the direction of the velocity, the fingers extended in the direction of the magnetic field and the palm in the direction of the force
Let's apply these relationships to this problem
.1) the magnitude of B
Let's use the proton data
B = F / q v sin θ
B = 2.10 10⁻¹⁶/ (1.6 10⁻¹⁹ 1.70 10³ sin 90)
B = 0.772 T
Let's perform the same calculation for the electron
B = 8.40 10⁻¹⁶ / (1.6 10⁻¹⁹ 4.70 10³ sin 90)
B = 1,117 T
We can see that there are two components of the magnetic field in the area
Let's look for the magnitude with the Pythagorean triangle
B² = Bₓ² + [tex]B_{y}[/tex]² + [tex]B_{z}[/tex]²
B = √(1,117² + 0.772²)
B = 1,358 T
2) let's use the right hand rule.
For the proton
The thumb in the positive direction of the x axis (+ x)
The palm in the positive direction of the y axis (+ y)
Therefore the extended fingers point in the negative direction of z axis (-z) this is the direction of the magnetic field
For the electron
The thumb in the negative direction of the z axis (-z)
Since the electron has a negative charge, the direction of the force is contrary to the direction of the palm, in the negative direction of the y-axis (-y)
Therefore the extended fingers point in the positive direction of the x-axis (+ x)
Let's build the magnetic field vector
B = Bₓ i ^ + [tex]B_{y}[/tex] j ^ + [tex]B_{z}[/tex] k ^
B = 1,117 i ^ + 0 j ^ - 0.772 k ^
The field address can be found by trigonometry
tan θ = Bₓ / [tex]B_{z}[/tex]
θ = tan⁻¹ (Bₓ / [tex]B_{z}[/tex])
θ = tan⁻¹ (1,117 / (- 0.772))
θ = -55.35°
3) the force for an electron
Another way to calculate the force is through determinants,
F= -1.6 10⁻¹⁶ [tex]\left[\begin{array}{ccc}i&j&k\\0&-3.70&0\\1.117&0&-0.772\end{array}\right][/tex]
F = -1.6 10-16 i (3.7 0.772) + k ^ (1,117 3.7)
F = (-4.57 10-16 i ^ - 6.61 10-16 k ^) N
The magnitude is
F = √ Fₓ² + [tex]F_{z}[/tex]²
F = √ (4.57² + 6.61²) 10⁻¹⁶
F = 8.04 10⁻¹⁶ N
The address is
tan θ = [tex]F_{z}[/tex] / [tex]F_{x}[/tex]
θ = tan⁻¹ (6.61 / 4.57)
θ = 55.34°
After applying Pythagorean's Theorem, the magnitude of the magnetic field is equal to 1.3578 Tesla.
Given the following data:
- Velocity of proton = 1.70 km/s = 1700 m/s.
- Force in y-direction (proton) = [tex]2.10\times10^{-16}\; N[/tex]
- Velocity of electron = 4.70 km/s = 4700 m/s.
- Force in z-direction (electron) = [tex]8.40\times10^{-16}\; N[/tex]
Scientific data:
- Charge of proton = [tex]1.67 \times 10^{-19} C[/tex].
- Charge of electron = [tex]1.67 \times 10^{-19} C[/tex].
How to determine the magnitude of the magnetic field.
Mathematically, the magnetic force in a magnetic field is given by this formula:
[tex]F=qVB\\\\B_p=\frac{F_p}{qV} \\\\B_p=\frac{2.10\times10^{-16}}{1.67 \times 10^{-19} \times 1700}\\\\B_p=0.772\;T[/tex]
For the electron:
[tex]F=qVB\\\\B_e=\frac{F_p}{qV} \\\\B_e=\frac{8.40\times10^{-16}}{1.67 \times 10^{-19} \times 4700}\\\\B_e=1.117\;T[/tex]
Next, we would determine the magnitude of the magnetic field by applying Pythagorean's Theorem:
[tex]B=\sqrt{B_e^2 + B_p^2} \\\\B=\sqrt{1.117^2 + 0.772^2}\\\\B=\sqrt{1.2477+0.5960} \\\\B=\sqrt{1.8437}[/tex]
B = 1.3578 Tesla.
How to determine the direction of the magnetic field (xz-plane).
[tex]Tan \theta = \frac{B_e}{B_p} \\\\\theta = tan ^{-1}( \frac{1.117}{0.772}) \\\\\theta =tan ^{-1}(1.4469)[/tex]
Angle = 55.35°.
How to calculate the magnitude of the magnetic force (−y-direction).
Mathematically, the magnetic force in a magnetic field is given by this formula:
[tex]F=qVB\\\\F=1.6 \times 10^{-19} \times 3700 \times 1.3578\\\\F = 8.04 \times 10^{-16}\;Newton[/tex]
In conclusion, the direction of this magnetic force (xz-plane) is negative in the z-plane and positive in x-direction.
Read more on magnetic field here: https://brainly.com/question/7802337
