An electron has a velocity of 6.0 × 10⁶ m/s in the positive x direction at a point where the magnetic field has the components B_x = 3.0 T, B_y = 1.5 T, and B_z = 2.0 T. What is the magnitude of the acceleration of the electron at this point? A) 2.1 × 10¹⁸ m/s² B) 1.6 × 10¹⁸ m/s² C) 2.6 × 10¹⁸ m/s² D) 3.2 × 10¹⁸ m/s² E) 3.7 × 10¹⁸ m/s²

Question

Quizplus · Accepted Answer

The magnetic force on an electron moving with velocity v in a magnetic field B is given by the formula F = qvB, where q is the charge of the electron. 
The acceleration of the electron is given by the formula a = F/m, where m is the mass of the electron. 
The magnitude of the acceleration of the electron is then given by |a| = |F|/m. 
Here, the velocity of the electron is given as v = 6.0×10^6 m/s in the positive x direction, and the magnetic field at the point is given as Bx = 3.0 T, By = 1.5 T, and Bz = 2.0 T. 
The magnetic field vector at the point is given as B = Bx i + By j + Bz k, where i, j, and k are the unit vectors in the x, y, and z directions respectively. 
The velocity vector of the electron is given as v = vx i, where vx is the component of the velocity in the x direction. 
Therefore, the magnetic force on the electron is given by the cross-product of v and B, which is given as F = qv x B. 
Using the vector identity (a x b) x c = (c . a)b - (c . b)a, we can find that F = qvB sin θ, where θ is the angle between v and B. 
Here, θ = 90° as the direction of v and B are perpendicular. 
Thus, F = qvB.
Substituting the given values, we get F = (1.6×10^-19 C)(6.0×10^6 m/s)(3.0 i + 1.5 j + 2.0 k) = 14.4×10^-13 i + 7.2×10^-13 j + 9.6×10^-13 k N.
The mass of the electron is given as 9.1×10^-31 kg. 
Therefore, the magnitude of the acceleration of the electron is given as |a| = |F|/m = sqrt( (14.4×10^-13)^2 + (7.2×10^-13)^2 + (9.6×10^-13)^2 ) / 9.1×10^-31 = 2.6×10^18 m/s^2. 
Therefore, the correct answer is (C) 2.6 × 10^18 m/s^2.

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