The charged particle moving in a uniform magnetic field of (3 i + 2 j ) T has an acceleration (4 i - x 2 j ) m/s ^2 . The value of x is __________.

Step-by-step solution

The magnetic force on a moving charged particle is given by F = q( v B ) . Since the acceleration a is in the direction of the magnetic force, it must be perpendicular to the magnetic field B . Therefore, the dot product of acceleration and magnetic field is zero: a B = 0 Substituting the given vectors: (4 i - x 2 j ) (3 i + 2 j ) = 0 (4)(3) + (- x 2 )(2) = 0 12 - x = 0 x = 12 Answer: 12