Question

An electron is moving moving with a velocity (2i + 3j) m/s an electric field of intensity (3i + 6j + 2k) V/m and a magnetic field of (2j + 3k) Tesla. Find the direction of Lorentz force acting on the electron from X-axis​A. theta = cos^(-1) 1/5^(1/2)B. theta = cos^(-1) 2/5^(1/2)C. theta = tan^(-1) 2/5^(1/2)D. theta = sin^(-1) 2/5^(1/2)

Answer 1

answer : option (B) cos^-1(2/√5)

Lorentz force is vector sum of electric and magnetic force.

i.e., F = qE + q(v × B)

where, q = 1.6 × 10^-19 C

E = (3i + 6j + 2k) V/m

B = (2j + 3k) T

v = (2i + 3j) m/s

so, F = q(3i + 6j + 2k) + q{(2i + 3j) × (2j + 3k)}

[ cross product of (2i + 3j) × (2j + 3k) = (9i - 6j + 4k)]

= (3qi + 6qj + 2qk) + q(9i - 6j + 4k)

= 12q i + 0j + 6q k

now direction of Lorentz force from x-axis, θ = tan^-1(y/x) = tan^-1(6q/12q) = tan^-1(1/2) or, sin^-1(1/√5) or, cos^-1(2/√5)

hence, option (B) is correct choice.

[ using basic trigonometric identities, if tanθ = 1/2 , then sinθ = 1/√5 and cosθ = 2/√5 ]

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Answer 2

$\huge\bold\purple{Answer:-}$

option (B) cos^-1(2/√5)