Question

A particle of charge -16× 10⁻¹⁸C moving with velocity 10ms⁻¹ along the x-axis entres a region where a magnetic field of induction B is along the y-axis and an electric field of magnetitude 10⁴ Vm⁻¹ is along the negative z-axis . If the charged particle continues moving along x-axis , the magnitude of B is [AIEEE 2003] ​

Answer 1

Answer:

The magnitude of B is 10³wb/m²

Explanation:

Given :- charge 'q' = -16× 10⁻¹⁸ C

velocity 'V'= 10ms⁻¹

Electric field 'E' = 10⁴ vm⁻¹

As we know that , F = q(E+v×B)

• or F = Fₑ + Fₘ

• ∴ Fₑ = qE = - 1.6× 10⁻¹⁸ × 10⁴ (-j)

• => 16 × 10⁻¹⁴k

• and Fₘ = -16 × 10⁻¹⁸ [ 10i + Bj]

• = -16 × 10⁻¹⁷ × B(+k)

• Fₘ = -16 × 10⁻¹⁷ Bk

since, Particle will continue to move along +x -axis , so resultant force is equal to zero

• Fₑ + Fₘ = 0
• Fₑ = -Fₘ

• => 16 × 10⁻¹⁴ = -(-16 × 10⁻¹⁷ B )

• => 16 × 10⁻¹⁴ /16 × 10⁻¹⁷ = 10³

• Or, B = 10³ wb/m² Answer