Help with physics question(magnetism)

Biala

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4 years ago

the details are in the attachment. Thank you again. I've been having a bit of trouble.

A proton moving with speed v = 2.6 x 10 m/s in a field-free region abruptly enters an essentially uniform magnetic field 0 850 T (B L)? If the proton enters the magnetic field region at a 45 angle as shown in the figure (Figure 1b)

At what angle does it leave

At what distance x does it exit the field?

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bio_man

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4 years ago

(a) Because the velocity is perpendicular to the magnetic field, the proton will travel in a circular arc. Then the force on the proton

F = q*V x B

From the symmetry of the motion we see that the upper half is a mirror image of the lower half, so the exit angle is the same as the incident angle: 45°.

(b) The magnetic force provides the radial acceleration, so we have F = evB = mv2/r, so r = mv / eB = (1.67 × 10–27 kg)(2.0 × 105 m/s)/(1.60 × 10–19 C)(0.850 T) = 2.46 × 10–3 m.　

Thus the distance x is 3.5 × 10–3 m.

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Biala Author

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4 years ago

Thank you Bio_man. It's good that we got a similar answer. I didn't want to use all my other tries. I posted another question too.

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