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An electron in a uniform electric and magnetic field has a velocity of 3.70x103 m/s in the positive x direction and an acceleration of 3.80x1012 m/s² in the positive z direction. If the electric field has a strength of 13.0 N/C (in the positive z direction), what is the MAGNITUDE of the magnetic field in the region?

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Final answer:

The magnitude of the magnetic field in the region is approximately 0.051 T.

Step-by-step explanation:

The velocity of the electron is given as 3.70x103 m/s in the positive x direction, and the acceleration is given as 3.80x1012 m/s² in the positive z direction. Since the electric field is in the positive z direction, it will not affect the electron's velocity in the x direction. Therefore, only the magnetic field is responsible for the acceleration in the z direction.

The force acting on the electron is given by F=qvB, where F is the force, q is the charge of the electron, v is the velocity of the electron, and B is the magnetic field strength. In this case, the force is equal to the mass of the electron (m) multiplied by the acceleration (a) in the z direction, F = ma.

Setting these two equations equal to each other, ma = qvB, we can solve for the magnetic field strength, B = ma / (qv).

Substituting the given values, B = (3.80x1012 m/s²) / ((-1.60x10-19 C)(3.70x103 m/s)), we find that the magnitude of the magnetic field is approximately 0.051 T.

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