When a charged particle moves at an angle of 11° with respect to a magnetic field, it experiences a magnetic force of magnitude F. At what angle (less than 90°) with respect to this field will this particle, moving at the same speed, experience a magnetic force of magnitude 2.4F?
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Verified answer
In absence of an electrical field the Lorentz force on the charged particle is given by :
F = q·(v × B) = (q·v) × B
(q charge , v velocity vector, B magnetic field vector)
The cross product can be expressed in terms of the magnitude of the two vectors, the angle θ between them
and the normal vector N perpendicular to the them:
F = |(q·v)|·|B| · sin θ · N
Hence the magnitude of the force is:
|F| = |(q·v)|·|B| · sinθ
For this problem
|F| = |(q·v)|·|B| · sin24°
|F'| = 2.4·|F| = |(q·v)|·|B| · sinθ'
|(q·v)| and |B| remain unchanged , thus
|F'| / |F| = sinθ' / sin24°
=>
θ' = arcsin(2.4·sin24°) = 77.5°