W.K.T, Magnetic Force qvB=mv^2/r ... n we know how to prove it too.... But how to show that the radius of curvature of the circular path is proportional to the linear momentum of the charge ?
To show that the radius of curvature of the circular path is proportional to the linear momentum of the charge, we can use the equation for centripetal force and the equation for linear momentum.
The equation for centripetal force is given by:
F = mv^2/r
Where:
F is the centripetal force,
m is the mass of the object,
v is the velocity of the object, and
r is the radius of curvature.
The equation for linear momentum is given by:
p = mv
Where:
p is the linear momentum of the object.
To relate the radius of curvature to the linear momentum, we can rearrange the centripetal force equation to solve for the radius:
r = mv^2 / F
Now, we can substitute the linear momentum (p) for the mass times velocity (mv) in the equation for the radius:
r = (p v) / F
Since p = mv, we can rewrite the equation as:
r = (p v) / F
Now, let's substitute the magnetic force equation for F (F = qvB):
r = (p v) / (qvB)
The charge (q) and the velocity (v) appear in both the numerator and the denominator, so they cancel out:
r = p/B
Therefore, we can conclude that the radius of curvature (r) of the circular path is directly proportional to the linear momentum (p) of the charge, with the constant of proportionality being 1/B.
So, the radius of curvature increases as the linear momentum of the charge increases, and vice versa.