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Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsElectric Field & ForceMedium2 minQB

PhysicsMediumsingle choice

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 $300 V / c m$ 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:

Options:

Answer:

Solution:

$q \lor B = q E$

$B = \frac{E}{V} = \frac{30000}{6 \times 10^{6}} = 5 \times 10^{3} T$

Magnetic force $F = q (\vec{V} \times \vec{B})$

So Ans. $5 \times 10^{3} T$ ,along +Z direction

Stream:JEESubject:PhysicsTopic:ElectrostaticsSubtopic:Electric Field & Force

⏱ 2mℹ️ Source: QB

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