Question

A proton is moving with a uniform velocity of \( 2 \times 10^8 \, \text{m/s} \) in uniform magnetic and electric fields which are perpendicular to each other. If the electric field is switched off, then the proton moves in a circular path of radius \( 1.6 \times 10^{-5} \, \text{m} \). Then the magnetic field \( B \) is:

• A. \( 5 \times 10^{-5} \, \text{T} \)
• B. \( 1.2 \times 10^5 \, \text{T} \)
• C. \( 2.5 \times 10^4 \, \text{T} \)
• D. \( 2.5 \times 10^2 \, \text{T} \)

Answer 1

The Correct Option is A

The force on a charged particle moving in a magnetic field is given by: \[ F = qvB = \frac{mv^2}{r}, \] where: - \( q \) is the charge of the proton (\( 1.6 \times 10^{-19} \, \text{C} \)), - \( v \) is the velocity of the proton (\( 2 \times 10^8 \, \text{m/s} \)), - \( B \) is the magnetic field, - \( m \) is the mass of the proton (\( 1.67 \times 10^{-27} \, \text{kg} \)), - \( r \) is the radius of the circular path (\( 1.6 \times 10^{-5} \, \text{m} \)). Rearranging the equation for \( B \): \[ B = \frac{mv}{qr}. \] Substitute the values: \[ B = \frac{(1.67 \times 10^{-27} \, \text{kg}) \times (2 \times 10^8 \, \text{m/s})}{(1.6 \times 10^{-19} \, \text{C}) \times (1.6 \times 10^{-5} \, \text{m})}. \] After calculating, we find \( B = 5 \times 10^{-5} \, \text{T} \). Thus, the correct answer is option (1).