Motion of a charged particle

Remember concepts with our Masterclasses.

80k Users

60 mins Expert Faculty Ask Questions

Question

What is the value of B (in $\times$ 10^-8 T ) that can be set up at the equator to permit a proton of speed $10^{7} m / s$ to circulate around the earth? $[R = 6.4 \times 10^{6} m, m_{P} = 1.67 \times 10^{- 27} kg]$

Moderate

Solution

$From the equation, r = \frac{m v}{B q} We have B = \frac{m v}{q r} S u b s t i t u t i n g t h e v a l u e s, w e h a v e B = \frac{(1.67 \times 10^{- 27}) (10^{7})}{(1.6 \times 10^{- 19}) (6.4 \times 10^{6})} = 1.6 \times 10^{- 8} T$

Ready to Test Your Skills?

Check Your Performance Today with our Free Mock Tests used by Toppers!

Talk to our academic expert!

Create Your Own Test
Your Topic, Your Difficulty, Your Pace

Similar Questions

The figure shows a circular region of radius $R = \sqrt{3} m$ which has a uniform magnetic field B=0.2 T directed into the plane of the figure. A particle having mass $m = 2 g$ , speed v=0.3m/s and charge q=1mC is projected along the radius of the circular region as shown in figure. Calculate the angular deviation (in degrees) produced in the path of the particle as it comes out of the magnetic field. Neglect any other force apart from the magnetic force.