Question

An electron (mass = \( 9 \times 10^{-31} \) kg, charge = \( 1.6 \times 10^{-19} \) C) moving with a velocity of \( 10^6 \) m/s enters a magnetic field. If it describes a circle of radius \( 0.1 \) m, then the strength of the magnetic field must be:

• A. \( 4.5 \times 10^{-5} \) T
• B. \( 1.4 \times 10^{-5} \) T
• C. \( 5.5 \times 10^{-5} \) T
• D. \( 2.6 \times 10^{-5} \) T

Hint:

For a charged particle in a magnetic field:
- The force acts perpendicular to velocity, causing circular motion.
- The radius of the path is given by \( r = \frac{m v}{q B} \).
- Higher velocity or mass increases the radius of curvature.

Answer 1

The Correct Option is C

Step 1: The force due to the magnetic field provides the centripetal force:
\[ q v B = \frac{m v^2}{r} \] Step 2: Rearranging for \( B \):
\[ B = \frac{m v}{q r} \] Step 3: Substituting given values:
\[ B = \frac{(9 \times 10^{-31}) (10^6)}{(1.6 \times 10^{-19}) (0.1)} \] Step 4: Simplifying:
\[ B = 5.5 \times 10^{-5} T \]