### Magnetic Force on a Proton in a Particle Accelerator#### Problem Statement:A proton in a particle accelerator has a speed of \( 5.0 \times 10^6 \) m/s. The proton encounters a magnetic field whose magnitude is 0.40 T and whose direction makes an angle of 30.0 degrees with respect to the proton’s velocity (see part (c) of the figure). Find (a) the magnitude and direction of the force on the proton and (b) the acceleration of the proton. (c) What would be the force and acceleration of the particle if it were an electron?#### Diagrams:The figure presented consists of three parts (a), (b), and (c), each depicting the being discussed in the problem:1. **Part (a)**: Shows the initial position of the proton as it enters the region of the magnetic field. - The proton is represented as a yellow sphere. - Arrows indicate the proton's initial velocity vector, \( \mathbf{v} \), in the positive x-direction.2. **Part (b)**: Illustrates the orientation and direction of the magnetic field. - The magnetic field vector, \( \mathbf{B} \), is represented as blue arrows. - The angle between the velocity vector \( \mathbf{v} \) of the proton and the magnetic field vector \( \mathbf{B} \) is labeled as 30.0 degrees.3. **Part (c)**: Combines the velocity and magnetic field vectors to illustrate their spatial relationship. - The proton's velocity \( \mathbf{v} \) is again shown in the positive x-direction. - The magnetic field vector \( \mathbf{B} \) is depicted making a 30.0-degree angle with respect to the velocity vector. - Coordinate axes are labeled to indicate directions.### Solution:#### (a) Magnitude and Direction of the Force on the Proton:The magnitude of the magnetic force \( \mathbf{F} \) experienced by the proton can be calculated using the formula:\[ F = qvB \sin(\theta) \]Where:- \( q \) is the charge of the proton (\( q = +1.6 \times 10^{-19} \) C).- \( v \) is the speed of the proton (\( v = 5.0 \times 10^6 \

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