Electrons traveling at a velocity of 2-4- 106ms-1 enter a region of cross electric and magnetic fields as shown in fig- If the electric field is 3-0- 106 Vm and the flux density of the magnetic field is 1-5 T- the electron upon entering the region of the crossed fields will

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Byju's Answer

Standard XII

Physics

The Circular Motion

Electrons tra...

Question

Electrons traveling at a velocity of $2.4 \times 10^{6} m s^{- 1}$ enter a region of cross electric and magnetic fields as shown in fig. If the electric field is $3.0 \times 10^{6} V m$ and the flux density of the magnetic field is 1.5 T, the electron upon entering the region of the crossed fields will

continue to travel undeflected in their original direction

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be deflected upward in the plane of the diagram

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be deflected downward on the plane of the diagram

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none of the above

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Solution

The correct option is C be deflected downward on the plane of the diagram
Electric force acting on particle is $q \to E$
$= e E$ upwards
$= 4.8 \times 10^{- 13} N$
Magnetic force acting on particle= $q (\to v \times \to B)$
$= e v B$ downwards
$= 5.76 \times 10^{- 13} N$
$F_{B} > F_{E}$
Hence particle moves downwards in the plane of diagram.

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Electrons traveling at a velocity of 2.4×106ms−1 enter a region of cross electric and magnetic fields as shown in fig. If the electric field is 3.0×106Vm and the flux density of the magnetic field is 1.5 T, the electron upon entering the region of the crossed fields will

The correct option is C be deflected downward on the plane of the diagramElectric force acting on particle is q→E=eE upwards=4.8×10−13NMagnetic force acting on particle=q(→v×→B)=evB downwards=5.76×10−13NFB>FEHence particle moves downwards in the plane of diagram.