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.

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