Question

A certain baryon (a subatomic particle) has a charge q > 0. The baryon moves with a kinetic energy K in a circular path of radius R in the presence of a uniform magnetic field of magnitude B. (Use any variable or symbol stated above as necessary.) What is the speed of the bayron? Assume the speed is low compared to the speed of light. What is the mass of the bayron?

Answer 1

Answer:

mass of the baryon = $\frac{(qRB)^{2} }{2K}$

speed of the baryon = $\frac{2K}{qRB}$

Explanation:

For any body to move in a circular path, there must be a centripetal force which is directed towards the center of the circle. Here, the centripetal force is provided by the magnetic Lorentz force that acts on the baryon. Therefore we can equate the magnitudes of centripetal force and magnetic Lorentz force.

Magnitude of Centripetal force = $\frac{mv^{2} }{R}$

Magnitude of Magnetic Lorentz force =  qvB

where,

m = mass of the baryon

v = velocity of the baryon

Thus,

$\frac{mv^{2} }{R}$ = qvB

$\frac{mv}{R}$ = $qB$ (cancelling v on both sides)

v = $\frac{qRB}{m}$

We know, Kinetic energy, K = $\frac{1}{2} mv^{2}$

Substituting v in the above equation we get,

K = $\frac{1}{2} m(\frac{qRB}{m} )^{2}$

K = $\frac{(qBR)^{2} }{2m}$ (simplifying)

Thus,

m = $\frac{(qRB)^{2} }{2K}$

We already got that

v = $\frac{qRB}{m}$

substituting the value of m in this equation gives,

$v = \frac{qRB}{\frac{(qRB)^{2} }{2K}}$

v = $\frac{2K}{qRB}$ (simplifying)

Thus,

mass of the baryon = $\frac{(qRB)^{2} }{2K}$

velocity of the baryon = $\frac{2K}{qRB}$