Jackals9912
24.06.2021 •
Physics
Un campo magn´etico uniforme B, con magnitud 1.2 mT, apunta verticalmente hacia arriba a lo largo del volumen del sal´on en que usted est´a sentado. Un prot´on de 5.3 MeV se mueve horizontalmente de sur a norte a trav´es de cierto punto en el salon. ¿Qu´e fuerza magn´etica deflectora act´ua sobre el prot´on cuando pasa por este punto? Cu´al es su aceleraci´on, la masa del prot´on es de 1.67x10−27 kg. Resp. 6.1x10−15 N; 3.7x1012 m/s2.
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Ответ:
F = 6.12 10⁻¹⁵ N, a = 3.66 10¹² m / s²
Explanation:
The magnetic force is
F = q v x B
bold letters indicate vectors, the modulus of this expression is
F = q v B sin θ
as they indicate that the magnetic field is vertical the proton moves horizontally the angle is 90º
suppose that the energy of the proton is totally kinetic
E = 5.3 MeV (10⁶ 1.6 10⁻¹⁹J / 1 eV) = 8.48 10⁻¹³ J
E = K = ½ m v²
v =
v =
v =
v = 3.19 10⁷ m / s
we calculate
F = 1.6 10⁻¹⁹ 3.19 10⁷ 1.2 10⁻³
F = 6.12 10⁻¹⁵ N
The direction of this out is given by the right hand rule,
thumb in the direction of speed, south to north
fingers extended in the direction of the magnetic field, vertical
the probe points in the direction of the force for a positive charge, East
To calculate the acceleration we use Newton's second law
F = ma
a = F / m
a = 6.12 10⁻¹⁵ / 1.67 10⁻²⁷
a = 3.66 10¹² m / s²
the direction of the acceleration is the same as the direction of the force
Ответ: