JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
An electron is moving along \(+x\) direction with a velocity of \(6 \times 10^{6}\, ms ^{-1}\). It enters a region of uniform electric field of \(300 \,V / cm\) pointing along \(+ y\) direction. The magnitude and direction of the magnetic field set up in this region such that the electron keeps moving along the \(x\) direction will be
- A \(5 \times 10^{-3} T ,\) along \(+ z\) direction
- B \(3 \times 10^{-4} T ,\) along \(- z\) direction
- C \(3 \times 10^{-4} T ,\) along \(+ z\) direction
- D \(5 \times 10^{-3} T ,\) along \(- z\) direction
Answer & Solution
Correct Answer
(A) \(5 \times 10^{-3} T ,\) along \(+ z\) direction
Step-by-step Solution
Detailed explanation
\(\overrightarrow{ B }\) must be in \(+ z\) axis. \(\overrightarrow{ V }=6 \times 10^{6} \hat{ i }\) \(\overrightarrow{ E }=300 \hat{ j } V / cm =3 \times 10^{4} V / m\) \(q \vec{ E }+q \vec{ V } \times \vec{ B }=0\) \(E = VB\)…
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