What magnitude electric field is necessary to exactly...

What magnitude electric field is necessary to exactly balance this magnetic force?
A. $1.8 \times 10^{-4} \mathrm{N} / \mathrm{C}$
?. $1.5 \times 10^{-4} \mathrm{N} / \mathrm{C}$
C. $1.0 \times 10^{-4} \mathrm{N} / \mathrm{C}$
D. $0.9 \times 10^{-4} \mathrm{N} / \mathrm{C}$

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Key Concepts

Magnetic Force

The magnetic force on a moving charge is given by the equation F = q(v × B), where q is the charge, v is the velocity, and B is the magnetic field. Its magnitude is qvB sin?, with ? being the angle between the velocity and the magnetic field vectors. When the angle is 90° (i.e., v is perpendicular to B), the equation simplifies to F = qvB.

Electric Force

The electric force acting on a charged particle in an electric field is described by F = qE, where q is the charge and E is the magnitude of the electric field. This force acts in the direction of the electric field if the charge is positive, or opposite to it if the charge is negative.

Equilibrium Condition

When a charged particle is subjected to both electric and magnetic forces, it can be brought to a state of equilibrium if these forces balance each other. For equilibrium, the magnitudes of the electric force and magnetic force must be equal, leading to the equation qE = qvB. From this, it follows that the necessary electric field magnitude to balance the magnetic force is E = vB.

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