The magnetic field in a region is given by
\[ B = (0.750 \, \hat{i} + 0.270 \, \hat{j}) \, \text{T}. \]

At some instant, a particle with charge
\[ q = 30.0 \, \text{mC} \]
has velocity
\[ v = (35.6 \, \hat{i} + 107.3 \, \hat{j} + 46.5 \, \hat{k}) \, \text{m/s}. \]

What is the magnetic force exerted on the particle at that instant? (Express your answer in vector form.)

The magnetic force exerted on the particle at that instant is given by the formula F = qvBsinθ, where θ is the angle between the velocity vector and the magnetic field vector.

Explanation:

To calculate the magnetic force exerted on the particle, we can use the formula:

F = qvBsinθ

Given:

Charge of the particle (q) = 30.0 mC
Velocity of the particle (v) = (35.6 + 107.3 + 46.5) m/s
Magnetic field (B) = (0.750 + 0.270) T

First, we need to calculate the angle (θ) between the velocity vector and the magnetic field vector. Since the question does not provide this information, we cannot determine the exact value of θ.

Next, we can substitute the given values into the formula:

F = (30.0 mC) * ((35.6 + 107.3 + 46.5) m/s) * ((0.750 + 0.270) T) * sin(θ)

Since the question asks for the answer in vector form, we need to include the direction of the force. The direction can be determined using the right-hand rule, which states that if you point your thumb in the direction of the velocity vector, and your fingers in the direction of the magnetic field vector, then the force vector will be perpendicular to both.

Therefore, the magnetic force exerted on the particle at that instant is given by the formula F = qvBsinθ, where θ is the angle between the velocity vector and the magnetic field vector.

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