Question

A charged particle projected in a magnetic field B=(3hati+4hatj)xx10^-2T The acceleration of the particle is found to be a=(xhati+2hatj)m/s^2 find the value of x

Answer 1

value of x would be -8/3 m/s²

given, a charged particle projected in a magnetic field as B = (3i + 4j) × 10^-2 T

acceleration of the particle is found to be a = (x i + 2 j) m/s².

we know, magnetic force is perpendicular on magnetic field.

so, acceleration due to magnetic force is also perpendicular on magnetic field.

i.e., dot product of a and B = 0

⇒a.B = (3i + 4j).(x i + 2j) = 0

⇒3x + 8 = 0

⇒x = -8/3

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