An electron moves with a speed of 5.0 × 104 m/s perpendicular to a uniform magnetic field of magnitude 0.20 T. What is the magnitude of the magnetic force on the electron? (e = 1.60 × 10-19 C)

F = 1.6 × 10⁻¹⁵ N

Explanation:

Given: V= 5.0 × 10⁴ m/s, B = 0.20 T, q = 1.60 × 10⁻¹⁹ C  and θ = 90°

To Find: F = ?

Solution:

we have a Formula for this question

F = q ( V x B)             ∴ x is used for cross product

F= q V B sin θ

F = 1.60 × 10⁻¹⁹ C × 5.0 × 10⁴ m/s × 0.20 T sin 90°

F = 1.6 × 10⁻¹⁵ N

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answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6