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NEET Physics QUESTION #1058
Question 1
An electron (mass \(9\times10^{-31}\ \text{kg}\), charge \(1.6\times10^{-19}\ \text{C}\)) moves at \(c/100\) in field \(B=9\times10^{-4}\ \text{T}\). Electric field \(E\) to prevent deflection:
  • E parallel to B, magnitude \(27\times10^2\ \text{V/m}\)
  • E parallel to B, magnitude \(27\times10^4\ \text{V/m}\)
  • E perpendicular to B, magnitude \(27\times10^4\ \text{V/m}\)✔️
  • E perpendicular to B, magnitude \(27\times10^2\ \text{V/m}\)
Correct Answer Explanation
For no deflection: electric force must balance magnetic force. \(qE=qvB\Rightarrow E=vB=\frac{3\times10^8}{100}\times9\times10^{-4}=3\times10^6\times9\times10^{-4}=27\times10^2\ \text{V/m}\). E must be perpendicular to B (to oppose magnetic force direction). So E perpendicular to B, magnitude \(27\times10^2\ \text{V/m}\) = option (4).