Respuesta :
Answer:
[tex]6.71 * 10^{14} m/s^{2}[/tex]
Explanation:
Parameters given;
Magnetic field intensity, B = 1.4 T
Speed of proton, v = [tex]5 * 10^{6} m/s[/tex]
Mass of proton, m = [tex]1.67 * 10^{-27} kg[/tex]
Charge of proton, q = [tex]1.6 * 10^{-19} C[/tex]
To find the acceleration of the proton, we first need to find the force exerted by the magnetic field on the proton:
F = q * v * B
F = [tex]1.6 * 10^{-19}[/tex] * [tex]5 * 10^{6}[/tex] * 1.4
F = [tex]1.12 * 10^{-12} N[/tex]
This is the force exerted by the magnetic field on the proton. The force exerted by the proton has the same magnitude but an opposite direction as the force exerted by the magnetic field. Hence, [tex]F_{p}[/tex] = - [tex]1.12 * 10^{-12} N[/tex]
The force exerted by the proton is the product of its mass and acceleration. Hence, we can find its acceleration:
[tex]F_p[/tex] = ma
a = F/m
[tex]a = \frac{-1.12 * 10^{-12} }{1.67 * 10^{-27}}[/tex]
[tex]a = -6.71 * 10^{14} m/s^{2}[/tex]
The magnitude, |a|, will be:
|a| =| [tex]-6.71 * 10^{14}[/tex]| [tex]m/s^2[/tex]
|a| = [tex]6.71 * 10^{14}[/tex] [tex]m/s^2[/tex]
The magnitude of the acceleration of the proton is |a| [tex]= 6.71 * 10^{14} m/s^{2}[/tex]
Answer:
The magnitude of the maximum acceleration of the proton is 6.707 x 10¹⁴ m/s²
Explanation:
Given;
magnitude of magnetic filed, B = 1.40 T
speed of the proton, v = 5.00 × 10⁶ m/s
mass of the proton, m = 1.67 × 10⁻²⁷ kg
charge of the proton, q = 1.6 × 10⁻¹⁹ C
Magnitude of force experienced by this proton due to the magnetic field is given as;
F = qvB
F = ( 1.6 × 10⁻¹⁹)(5.00 × 10⁶ )(1.40)
F = 1.12 x 10⁻¹² N
The magnitude of the maximum acceleration of the proton:
From Newton's second law of motion;
F = ma
a = F/m
a = (1.12 x 10⁻¹²) / (1.67 × 10⁻²⁷)
a = 6.707 x 10¹⁴ m/s²
Thus, the magnitude of the maximum acceleration of the proton is 6.707 x 10¹⁴ m/s²
