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  • I need help with - Magnetic Effects of Current and Magnetism - JEE Main

A charged particle with charge  q enters a region of constant, uniform and mutually orthogonal fields \vec{E}\; and \; \vec{B}  with a velocity

\vec{v} perpendicular to both \vec{E}\; and \; \vec{B} , and comes out without any change in magnitude or direction of \vec{v}. Then

  • Option 1)

    \vec{v}=\vec{B}\times \vec{E}/E^{2}

  • Option 2)

    \vec{v}=\vec{E}\times \vec{B}/B^{2}

  • Option 3)

    \vec{v}=\vec{B}\times \vec{E}/B^{2}

  • Option 4)

    \vec{v}=\vec{E}\times \vec{B}/E^{2}

 

Answers (1)

As we learnt in

Magnetic field If V(vector), E (vector) and B (vector) are mutually perpendicular -

Fe=Fm

V=\frac{E}{B}

-

 

 

 

When \vec{E}\; and \; \vec{B} are perpendicular and velocity has no changes then qE=qvB\, \, i.e.,\: v=\frac{E}{B}.  The two forces oppose each other if v  is along  \vec{E}\times \vec{B}  i.e.,

\vec{v}=\frac{\vec{E}\times \vec{B}}{B^{2}}

As \vec{E}\; and \; \vec{B}   are perpendicular to each other

\frac{\vec{E}\times \vec{B}}{B^{2}}=\frac{EB\, sin\, 90^{0}}{B^{2}}=\frac{E}{B}

 


Option 1)

\vec{v}=\vec{B}\times \vec{E}/E^{2}

Incorrect option

Option 2)

\vec{v}=\vec{E}\times \vec{B}/B^{2}

Correct option

Option 3)

\vec{v}=\vec{B}\times \vec{E}/B^{2}

Incorrect option

Option 4)

\vec{v}=\vec{E}\times \vec{B}/E^{2}

Incorrect option

Posted by

Vakul

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