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What would be the magnetic force on a hypothetical particle of charge 1.0 × 10⁻¹⁹C moving with a velocity of 6.0 × 10⁴m/s i in a magnetic field

B=(0.4i+1.2k) T?
Show all calculations and include the direction of the force vector using (i,j,k) notation.

1 Answer

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Final answer:

The magnetic force on the particle is -7.2 × 10⁻¹⁵ N in the negative j-direction (downwards). This is found by calculating the cross product of the given velocity vector and magnetic field vector.

Step-by-step explanation:

The magnetic force on a charged particle moving in a magnetic field can be calculated using the Lorentz force equation F = q(v x B), where q is the charge, v is the velocity vector and B is the magnetic field vector.

The charge of the particle is 1.0 × 10⁻¹⁹C, the velocity is given as 6.0 × 10⁴m/s i, and the magnetic field is B=(0.4i+1.2k) T. To find the magnetic force, we use the cross product of the velocity and magnetic field vectors:

F = q(v x B) = (1.0 × 10⁻¹⁹C)((6.0 × 10⁴ mi) x (0.4mi + 1.2kk))

Calculating the cross product:

mi x mi = 0, and mi x kk = -mj

F = (1.0 × 10⁻¹⁹C)((6.0 × 10⁴)(0) mi + (6.0 × 10⁴)(1.2)(-mj))

F = -7.2 × 10⁻¹⁵ N j

The direction of the force vector is in the negative j-direction, which means it is pointing downwards when using the right-hand rule with conventional current direction (positive charge).

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