A proton of energy `200 MeV` enters the magnetic field of `5 T`. If direction of field is from south to north and motion is upward, the force acting on it will be

Explanation

Because the proton is charged and moving through a magnetic field, a magnetic force `vec(F)_(B)` can act on it. Because the initial direction of the proton velocity is not along a magnetic field line, `vec(F)_(B)` is not simply zero.
Magnitude : To find the magnitude of `vec(F)_(B)`, we can use Eq. 28-3 `(F_(B) - |q|v B sin phi)` provided we first find the proton.s speed v. We can find v from the given kinetic energy because `K = 1//2 mv^(2)`, Solving for v, we obtain
`v = sqrt((2K)/(m)) = sqrt(((2)(5.3 MeV)(1.60 xx 10^(-13)J//MeV))/(1.67 xx 10^(-27)kg))`
`= 3.2 xx 10^(-7) m//s`
Equation 28-3 then yields
`F_(B) = |q|v B sin phi`
`= (1.60 xx 10^(-19) C)(3.2 xx 10^(7) m//s) xx (1.2 xx 10^(-3)T) (sin 90^(@))`
`= 6.1 xx 10^(-15)N`.
This may seem like a small force, but it acts on a particle of small mass, producing a large acceleration, namely,
`a = (F_(B))/(m) = (6.1 xx 10^(-15)N)/(1.67 xx 10^(-27) kg) = 3.7 xx 10^(12) m//s^(2)`.
Direction : To find the direction of `vec(F)_(B)`, we use the fact that `vec(F)_(B)` has the direction of the cross product `q vec(v) xx vec(B)`. Because the charge q is positiv , `vec(F)_(B)` must have the same direction as `vec(v) xx vec(B)`, which can determined with the right-hand rule for cross products (as in Fig.) We know that `vec(v)` is directed horizontally from south to north and `vec(B)` is directed vertically up. The right-hand rule shows us that the deflecting force `vec(F)_(B)` must be directed horizontally from west to east, as Fig. shows.
If the charge of the particle were negative, the magnetic deflecting force would be directed in the opposite direction - that is, horizontally from east to west. This is predicted automatically by Eq. 28-2 if we substitute a negative value for q.

An downward view of a proton moving from south to north with velocity `vec(v)` in a chamber. A magnetic field is directed vertically upward in the chamber, as represented by the array of dots (which resemble the tips of arrows). The proton is deflected toward the east.