JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2020
- Hard
- MCQ
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
Pick an option to check your answer.
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.
\(\overrightarrow{ V }=6 \times 10^{6} \hat{ i }\)
\(\overrightarrow{ E }=300 \hat{ j } V / cm =3 \times 10^{4} V / m\)
\(q \vec{ E }+q \vec{ V } \times \vec{ B }=0\)
\(E = VB\)
\(B =\frac{ E }{ V }=\frac{3 \times 10^{4}}{6 \times 10^{6}}=5 \times 10^{-3} T\)
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