Question 295: Magnetism and Matter

Question

A singly ionized magnesium atom (A = 24) ion is accelerated to kinetic energy 5 keV, and is projected perpendicularly into a magnetic field B of the magnitude 0.5 T. The radius of path formed will be _____________ cm.

Accepted Answer

To calculate the radius of the path formed by a singly ionized magnesium atom in a magnetic field, the following formula is used:
$ R = \frac{mv}{qB} $
This can be rewritten using kinetic energy (KE):
$ R = \frac{\sqrt{2m \cdot KE}}{qB} $
Here's how we calculate it for a magnesium ion:
Atomic Mass (A)
: 24
Mass of a Nucleon
: $1.67 	imes 10^{-27} $ kg (approximate mass of a proton or neutron)
Charge (q)
: $1.6 	imes 10^{-19} $ C (since the magnesium ion is singly ionized)
Magnetic Field (B)
: 0.5 T
Kinetic Energy (KE)
: 5 keV = $5 	imes 1.6 	imes 10^{-16} $ J
Plug in these values to find the radius:
$ R = \frac{\sqrt{2 	imes 24 	imes 1.67 	imes 10^{-27} 	imes 5 	imes 1.6 	imes 10^{-16}}}{1.6 	imes 10^{-19} 	imes 0.5} $
This simplifies to:
$ R = 10.009 \, 	ext{cm} \approx 10 \, 	ext{cm} $
Thus, the radius of the path is approximately 10 cm.

Magnetism and Matter

A singly ionized magnesium atom (A = 24) ion is accelerated to kinetic energy 5 keV, and is projected perpendicularly into a magnetic field B of the magnitude 0.5 T. The radius of path formed will be _____________ cm.

Choose the correct answer: