Question

An electron moving along a straight line with a velocity of $6.4 \times 10^7 ms^{-1}$ is subjected to a magnetic field of 3 mT. If the magnetic field is applied perpendicular to the direction of the initial path of the electron, then the radius of the circular path in which the electron moves under the influence of the magnetic field is (Mass of electron $= 9 \times 10^{-31}$ kg and charge of the electron $= 1.6 \times 10^{-19}$ C)

• A. 30 cm
• B. 12 cm
• C. 24 cm
• D. 36 cm

Hint:

Centripetal force is provided by magnetic Lorentz force: $\frac{mv^2}{r} = qvB \Rightarrow r = \frac{mv}{qB}$.

Answer 1

The Correct Option is B

Step 1: Key Formula:
The radius of the circular path of a charged particle in a perpendicular magnetic field is given by: \[ r = \frac{mv}{qB} \]
Step 2: Substitution:
$m = 9 \times 10^{-31}$ kg. $v = 6.4 \times 10^7 \, ms^{-1}$. $q = 1.6 \times 10^{-19}$ C. $B = 3 \, mT = 3 \times 10^{-3}$ T. \[ r = \frac{(9 \times 10^{-31}) \times (6.4 \times 10^7)}{(1.6 \times 10^{-19}) \times (3 \times 10^{-3})} \]
Step 3: Calculation:
Group numbers and powers: \[ r = \left( \frac{9 \times 6.4}{1.6 \times 3} \right) \times 10^{-31 + 7 - (-19) - (-3)} \] Numbers: $\frac{9}{3} \times \frac{6.4}{1.6} = 3 \times 4 = 12$. Powers: $-31 + 7 + 19 + 3 = -2$. \[ r = 12 \times 10^{-2} \, m = 12 \, cm \]
Step 4: Final Answer:
The radius is 12 cm.