Magnetic force:

$\overline{)\stackrel{\mathbf{\rightharpoonup}}{\mathbf{F}}{\mathbf{=}}{\mathbf{q}}\stackrel{\mathbf{\rightharpoonup}}{\mathbf{v}}\stackrel{\mathbf{\rightharpoonup}}{\mathbf{B}}{\mathbf{s}}{\mathbf{i}}{\mathbf{n}}{\mathbf{\theta}}}$

**(a)**

q = 1.6 × 10^{-19} C

v = 1.0 × 10^{7}m/s

B = 0.40 T

θ = 45°

The field is directed in the positive x-direction. The velocity is directed in the positive z-direction making an angle of 45° with the field. Thus, the resultant force will be in the y-direction.

F = (1.6 × 10^{-19})(1.0 × 10^{7})(0.40) sin 45 = 4.5 × 10^{-13} N j

A proton moves in the magnetic field B? =0.40i^T with a speed of 1.0 x 10^{7}m/s in the directions shown in the figure.

-In Figure (a), what is the magnetic force **F** on the proton? Give your answers in component form.

F = ??? N

-In Figure (b), what is the magnetic force **F **on the proton? Give your answers in component form.

F = ??? N

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