NEET · Physics · STD 12 - 4. Moving charges and magnetism
An electron (mass \(9 \times 10^{-31} kg\) and charge \(1.6 \times 10^{-19} C\) ) moving with speed \(c / 100( c =\) speed of light) is injected into a magnetic field \(\vec{B}\) of magnitude \(9 \times 10^{-4} T\) perpendicular to its direction of motion. We wish to apply an uniform electric field \(\vec{E}\) together with the magnetic field so that the electron does not deflect from its path. Then (speed of light \(c =3 \times 10^8 ms^{-1}\) )
- A \(\vec{E}\) is perpendicular to \(\vec{B}\) and its magnitude is \(27 \times 10^4 V m ^{-1}\)
- B \(\overrightarrow{ E }\) is parallel to \(\overrightarrow{ B }\) and its magnitude is \(27 \times 10^2 V m ^{-1}\)
- C \(\overrightarrow{ E }\) is perpendicular to \(\overrightarrow{ B }\) and its magnitude is \(27 \times 10^2 Vm ^{-1}\)
- D \(\overrightarrow{ E }\) is parallel to \(\overrightarrow{ B }\) and its magnitude is \(27 \times 10^4 V m ^{-1}\)
Answer & Solution
Correct Answer
(C) \(\overrightarrow{ E }\) is perpendicular to \(\overrightarrow{ B }\) and its magnitude is \(27 \times 10^2 Vm ^{-1}\)
Step-by-step Solution
Detailed explanation
For particle To Remain undeflected
\(\overrightarrow{ F }_e=-\overrightarrow{ F }_{ m } \)
\( \Rightarrow|\overrightarrow{ E }|= VB \sin \theta \)
\( E =\frac{ c }{100} \times 9 \times 10^{-4} \sin 90 \)
\( E =27 \times 10^2 V / m \text { and } \)
\( \overrightarrow{ E }=\overrightarrow{ B } \times \overrightarrow{ V}\)
\(\overrightarrow{ F }_e=-\overrightarrow{ F }_{ m } \)
\( \Rightarrow|\overrightarrow{ E }|= VB \sin \theta \)
\( E =\frac{ c }{100} \times 9 \times 10^{-4} \sin 90 \)
\( E =27 \times 10^2 V / m \text { and } \)
\( \overrightarrow{ E }=\overrightarrow{ B } \times \overrightarrow{ V}\)
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