Question

An electron having kinetic energy of \(100\,\text{eV}\) circulates in a path of radius \(10\,\text{cm}\) in a magnetic field. The magnitude of magnetic field \(|\vec{B}|\) is approximately \([\text{Mass of electron}=0.5\,\text{MeV}\,C^{-2},\text{ where }C\text{ is the velocity of light}]\).

• A. \(3.3\times10^{-4}\,\text{T}\)
• B. \(2.6\times10^{-4}\,\text{T}\)
• C. \(1.70\times10^{-4}\,\text{T}\)
• D. \(4.3\times10^{-4}\,\text{T}\)

Hint:

For a charged particle moving in a circular path inside a magnetic field, \[ B=\frac{p}{qr} \] and for non-relativistic kinetic energy, \[ pc=\sqrt{2mc^2K} \] is often useful when mass is given in \(\text{MeV}/c^2\).

Answer 1

The Correct Option is A

Step 1: Use magnetic force as centripetal force.
For an electron moving in a circular path in a magnetic field, \[ qvB=\frac{mv^2}{r} \] Therefore, \[ B=\frac{mv}{qr} \] Since, \[ p=mv, \] we can write \[ B=\frac{p}{qr} \]

Step 2: Find momentum using kinetic energy.
For non-relativistic motion, \[ K=\frac{p^2}{2m} \] So, \[ p=\sqrt{2mK} \] Multiplying by \(c\), \[ pc=\sqrt{2mc^2K} \] Given, \[ mc^2=0.5\,\text{MeV}=5\times10^5\,\text{eV} \] and \[ K=100\,\text{eV} \] Thus, \[ pc=\sqrt{2\times 5\times10^5\times 100} \] \[ pc=\sqrt{10^8} \] \[ pc=10^4\,\text{eV} \] Hence, \[ p=\frac{10^4\,\text{eV}}{c} \]

Step 3: Convert momentum into SI units.
\[ p=\frac{10^4\times 1.6\times10^{-19}}{3\times10^8} \] \[ p=5.33\times10^{-24}\,\text{kg m s}^{-1} \]

Step 4: Calculate magnetic field.
Given, \[ q=1.6\times10^{-19}\,\text{C} \] and \[ r=10\,\text{cm}=0.1\,\text{m} \] Using \[ B=\frac{p}{qr} \] \[ B=\frac{5.33\times10^{-24}}{1.6\times10^{-19}\times0.1} \] \[ B=3.33\times10^{-4}\,\text{T} \]

Step 5: Final conclusion.
Hence, the magnetic field is approximately \[ \boxed{3.3\times10^{-4}\,\text{T}} \]