WILL RATE A proton with speed of 1 Times 107 m/s enters a region of uniform magn

Question

WILL RATE

A proton with speed of 1 Times 107 m/s enters a region of uniform magnetic field of magnitude 1 T, directed into the page, as shown in the figure. The proton enters the magnetic field at an angle of 45 degree with respect to the normal of the fields boundary and exits the magnetic field at an angle phi with respect to the normal of the fields boundary at a distance d from the place where the charged particle entered the magnetic field. The charge on electron is 1.60218 Times 10-19 C. Figure: Not drawn to scale. Find the angle phi. Answer in units of degree The charge on a proton is 1.60218 Times 10-19 C and its mass is 1.67262 Times 10-27 kg. Find the distance d.

Explanation / Answer

The angle is 45 degrees and the distance d is 0.147639 m.

To see this, consider the geometry. The proton in the magnetic field will travel in a circular path, and the side of the magnetic field constitutes a chord of the circle. The tangent to the circle is the same at both ends of the chord, therefore the angle is 45 degrees.

The calculation of d is simple because of the 45 degree angle, which means that the circular path within the field is equal to exactly one quarter of a circl (because the tangents going in and coming out are perpendicular). We need the radius of the circle:

r = qv/mB = (1.60218 * 10-19 C)(1 * 107 m/s)/((1.67262 * 10-27 kg)(1 T) = 0.104397 m.

The distance d is equal to this number times the square root of 2, or 0.147639 m.

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