A particle of mass 10.6 g and charge [tex]99.9 \mu C[/tex] moves through a uniform magnetic field in a region where the free-fall acceleration is [tex]-9.8 \, \mathbf{j} \, \text{m/s}^2[/tex] without falling. The velocity of the particle is a constant [tex]24.1 \, \mathbf{i} \, \text{km/s}[/tex], which is perpendicular to the magnetic field. What, then, is the magnetic field?

Provide your answer as a vector in the form [tex](i^+ \mid j^+ \mid k^)[/tex] and include units.

moving in a magnetic field is given by the equation F = qvB, where F is the force, q is the charge, v is the velocity, and B is the magnetic field.

In this case, the particle has a mass of 10.6 g, which is equivalent to 0.0106 kg, and a charge of 99.9 μC, which is equivalent to 99.9 × 10^(-6) C. The velocity of the particle is given as 24.1i^ km/s, where i^ represents the x-direction.

Since the velocity is perpendicular to the magnetic field, the force experienced by the particle is solely due to the magnetic field. Therefore, the magnetic force is equal to the centripetal force, which is given by F = m * a, where m is the mass and a is the centripetal acceleration.

By equating these two forces and solving for the magnetic field B, we find that B = m * a / (q * v). Substituting the given values, we calculate the magnetic field to be approximately 0.412 T in the positive k^ direction.

In conclusion, the magnetic field in the region where the particle is moving with a constant velocity of 24.1i^ km/s and experiencing a free-fall acceleration of -9.8j^ m/s^2 is approximately 0.412 T in the positive k^ direction. This calculation is based on the equation for the magnetic force and the given values of mass, charge, and velocity.

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