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Question: A proton moves at a speed of 1.9 × 10^7 m/s at right angles to a magnetic field with a magnitude of 0.49 T. Find the magnitude of the acceleration of the proton. The elemental charge is 1.60 × 10^−19 C .

The force on a moving charged particle in a magnetic field is given by the formula F=Bqv, where B is the magnetic field strength, q is the charge, and v is the velocity. The force is orthogonal to the velocity of the particle and the direction of the field, which leads to the particle's centripetal acceleration in a circular path. In your case, knowing that centripetal force is equal to the mass (m) of the particle times its acceleration (a), we can set up the equation: F=ma. Substituting m*a with Bqv (provided by the Lorentz force law) gives: Bqv = ma. We are looking for the acceleration (a), so let's rearrange the equation for a: a= Bqv/m. Here is what we know: - B (the magnetic field strength) is 0.49 T. - q (the charge) is 1.60 × 10^-19 C. - v (the velocity) is 1.9 × 10^7 m/s. - m (the mass of the proton) is around1.67 × 10^-27 kg. Substituting these values into the formula we get: a=(0.49 T * 1.60 × 10^-19 C * 1.9 × 10^7 m/s) / 1.67 × 10^-27 kg. By calculating this, the answer for the acceleration will be approximately 8.83 × 10^13 m/s². Please keep in mind that the answer is given in two significant digits following the scientific notation rules.

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