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NEET2026PhysicsChapterActual

A particle having a charge of -2 C is moving with a velocity v = (v_x i + 3 j ) m/s in a uniform magnetic field B = (2 i - 4 j + B_z k ) T . If the magnetic force acting on the particle is F = (24 i + 16 j - 4 k ) N , what are the values of v_x and B_z ?

Options

  1. Av_x = -2 m/s , ; B_z = 4 T
  2. Bv_x = 2 m/s , ; B_z = -4 T
  3. Cv_x = -2 m/s , ; B_z = -4 T
  4. Dv_x = 4 m/s , ; B_z = 2 T

Correct answer

C. v_x = -2 m/s , ; B_z = -4 T

Step-by-step solution

The magnetic force on a moving charge is given by F = q( v B ) . Substituting the given vectors: v B = (v_x i + 3 j ) (2 i - 4 j + B_z k ) v B = 3 B_z i - v_x B_z j + (-4 v_x - 6) k Multiplying by the charge q = -2 C : F = -2 [3 B_z i - v_x B_z j + (-4 v_x - 6) k ] F = -6 B_z i + 2 v_x B_z j + (8 v_x + 12) k Equating this to the given force F = 24 i + 16 j - 4 k : For the i component: -6 B_z = 24 B_z = -4 T For the j component: 2 v_x B_z = 16 Substituting B_z = -4 T : 2 v_x (-4) = 16 -8 v_x = 16 v_x = -2 m/s Checki

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