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

Concept: When a charged particle enters a uniform magnetic field at an angle other than $0^\circ$ or $90^\circ$, its velocity can be resolved into two mutually perpendicular components: \[ v_{\parallel}=v\cos\theta \] along the magnetic field and \[ v_{\perp}=v\sin\theta \] perpendicular to the magnetic field. The magnetic force acts only on the perpendicular component of velocity. Therefore:
• The parallel component remains unchanged because no magnetic force acts along the field direction.
• The perpendicular component produces uniform circular motion.
• The combination of uniform circular motion and uniform linear motion results in a helical path. The pitch of the helix is defined as the distance travelled by the particle parallel to the magnetic field during one complete revolution. Hence, \[ \text{Pitch }(p)=v_{\parallel}T, \] where $T$ is the time period of circular motion. For a charged particle moving in a magnetic field, \[ T=\frac{2\pi m}{qB}. \] Therefore, \[ p=v\cos\theta\left(\frac{2\pi m}{qB}\right). \]

Step 1: Write down all the given quantities in SI units. Given, \[ v=4\times10^6\text{ ms}^{-1}, \] \[ \theta=60^\circ, \] \[ B=\frac{\pi}{2}\text{ mT} =\frac{\pi}{2}\times10^{-3}\text{ T}, \] \[ m=9\times10^{-31}\text{ kg}, \] and \[ q=1.6\times10^{-19}\text{ C}. \] All quantities are already in SI units.

Step 2: Calculate the time period of revolution of the electron. Using \[ T=\frac{2\pi m}{qB}, \] we get \[ T= \frac{2\pi(9\times10^{-31})} {(1.6\times10^{-19}) \left(\frac{\pi}{2}\times10^{-3}\right)}. \] Substituting the values, \[ T= \frac{18\pi\times10^{-31}} {0.8\pi\times10^{-22}}. \] The factor $\pi$ cancels from numerator and denominator: \[ T= \frac{18\times10^{-31}} {0.8\times10^{-22}}. \] Therefore, \[ T= \frac{18}{0.8}\times10^{-9}. \] \[ T=22.5\times10^{-9}\text{ s}. \] Thus, the electron completes one revolution in \[ \boxed{T=22.5\times10^{-9}\text{ s}}. \]

Step 3: Determine the component of velocity parallel to the magnetic field. The component parallel to the magnetic field is \[ v_{\parallel} = v\cos\theta. \] Substituting the given values, \[ v_{\parallel} = 4\times10^6\cos60^\circ. \] Since \[ \cos60^\circ=\frac{1}{2}, \] we obtain \[ v_{\parallel} = 4\times10^6\times\frac12. \] Hence, \[ v_{\parallel} = 2\times10^6\text{ ms}^{-1}. \]

Step 4: Calculate the pitch of the helical path. Pitch is given by \[ p=v_{\parallel}T. \] Substituting the values obtained above, \[ p= (2\times10^6) (22.5\times10^{-9}). \] Multiplying the numerical values, \[ p= 45\times10^{-3}\text{ m}. \] Thus, \[ p=0.045\text{ m}. \]

Step 5: Convert the answer into centimetres. Since \[ 1\text{ m}=100\text{ cm}, \] we have \[ p=0.045\times100. \] Therefore, \[ p=4.5\text{ cm}. \] Final Conclusion: The pitch of the helical path followed by the electron is \[ \boxed{4.5\text{ cm}}. \] Hence, the correct answer is \[ \boxed{\text{(C)}\;4.5\text{ cm}}. \]