A charged particle moves with an initial velocity v = (2 i + 3 j - k ) m s ⁻¹ in a uniform magnetic field B . It experiences an acceleration a = (x i + 2 j + 4 k ) m s ⁻² . Which of the following vectors could represent the uniform magnetic field B ?

Step-by-step solution

The magnetic force on a moving charge is F = q( v B ) . This implies that the force, and hence the acceleration a , is perpendicular to both the velocity v and the magnetic field B . First, using a v , we have a v = 0 : (x i + 2 j + 4 k ) (2 i + 3 j - k ) = 0 2x + 6 - 4 = 0 2x + 2 = 0 x = -1 So, the acceleration vector is a = - i + 2 j + 4 k . Second, using a B , we must have a B = 0 . We check the given options: For 4 i + 2 j : (- i + 2 j + 4 k ) (4 i + 2 j ) = -4 + 4 + 0 = 0 . This is a valid magnetic field. Note