Question

A velocity selector is to be constructed to select ions with a velocity of 6 km/s. If the electric field used is 400 V/m, then the magnetic field to be used is

• A. \(\frac{11}{20} \, T\)
• B. \(\frac{2}{3} \, T\)
• C. \(\frac{1}{15} \, T\)
• D. None of the above

Hint:

In a velocity selector, use \( B = \frac{E}{v} \) to find the magnetic field for given electric field and velocity.

Answer 1

The Correct Option is C

Step 1: Principle of velocity selector
In a velocity selector, electric force equals magnetic force: \[ q E = q v B \implies B = \frac{E}{v} \] Step 2: Substitute values
\[ E = 400\, V/m, \quad v = 6\, km/s = 6000\, m/s \] \[ B = \frac{400}{6000} = \frac{2}{30} = \frac{1}{15} \, T \] Step 3: Conclusion
The magnetic field required is \(\frac{1}{15} \, T\).