Question

If an electron moving with a velocity of $4 \times 10^6\text{ ms}^{-1}$ enters a uniform magnetic field of $\frac{\pi}{2}\text{ mT}$ at an angle of $60^\circ$ with the direction of the magnetic field, then the pitch of the helical path of the electron is: (Mass of the electron $= 9 \times 10^{-31}\text{ kg}$)

• A. $1.5\text{ cm}$
• B. $3\text{ cm}$
• C. $4.5\text{ cm}$
• D. $6\text{ cm}$

Hint:

Whenever a charged particle enters a magnetic field at an angle $\theta$, immediately resolve the velocity into \[ v_{\parallel}=v\cos\theta \] and \[ v_{\perp}=v\sin\theta. \] The perpendicular component determines the circular motion, while the parallel component determines the pitch. For helical motion, always use \[ p=v_{\parallel}T \] with \[ T=\frac{2\pi m}{qB}. \] If the magnetic field contains a factor of $\pi$, it often cancels directly with the $\pi$ present in the time-period formula, making calculations much faster.

Answer 1

The Correct Option is C

Step 1: Split the velocity.
The electron enters the field at $60^\circ$, so its velocity has a part along the field, $v_\parallel = v\cos\theta$, that stays uniform, and a part across the field, $v_\perp = v\sin\theta$, that makes it circle. The result is a helix.
Step 2: Recall the pitch.
The pitch is the forward distance covered in one full revolution, $p = v_\parallel T$, where the period is $T = \dfrac{2\pi m}{qB}$.
Step 3: List the data.
$v = 4\times 10^6\,ms^{-1}$, $\theta = 60^\circ$, $B = \dfrac{\pi}{2}\times 10^{-3}\,T$, $m = 9\times 10^{-31}\,kg$, $q = 1.6\times 10^{-19}\,C$.
Step 4: Compute the period.
$T = \dfrac{2\pi (9\times 10^{-31})}{(1.6\times 10^{-19})(\tfrac{\pi}{2}\times 10^{-3})}$. The $\pi$ cancels: $T = \dfrac{2(9\times 10^{-31})\times 2}{1.6\times 10^{-19}\times 10^{-3}} = \dfrac{36\times 10^{-31}}{1.6\times 10^{-22}} = 2.25\times 10^{-8}\,s$.
Step 5: Compute the pitch.
$v_\parallel = 4\times 10^6\times \cos 60^\circ = 2\times 10^6\,ms^{-1}$. So $p = (2\times 10^6)(2.25\times 10^{-8}) = 4.5\times 10^{-2}\,m = 4.5\,cm$.
Step 6: Conclude.
The pitch of the helical path is $4.5\,cm$, which is option (3).
\[ \boxed{p = 4.5\,cm} \]