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PhysicsMediumMCQ2020 · 06 Sep Shift 1

Q13.An electron is moving along +x direction with a velocity of 6 × 106 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: (1) 3 × 10−4T, along +z direction (2) 5 × 10−3T, along −z direction (3) 5 × 10−3T, along +z direction (4) 3 × 10−4T, along −zdirection JEE Main 2020 (06 Sep Shift 1) JEE Main Previous Year Paper

What This Question Tests

This problem is a classic application of the velocity selector principle, where electric and magnetic forces on a charged particle balance each other, requiring vector analysis to find the magnetic field's direction.

Concepts Tested

Lorentz forceElectric forceMagnetic forceCross product of vectors

Formulas Used

F_e = qE

F_m = q(v x B)

F_e + F_m = 0

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