Motion in a Magnetic Field 2 1 2 3 4 5 6 A proton (q 1.6X 1019 C, m-1.67 X 1027

Question

Motion in a Magnetic Field 2 1 2 3 4 5 6 A proton (q 1.6X 1019 C, m-1.67 X 1027 kg) moving with constant velocity enters a region containing a constant magnetic field that is directed along the z axis at (x,y)-(0,0) as shown. The magnetic field extends for a distance D 0.5 m in the x-direction. The proton leaves the field having a velocity vector (o (4.1 X 105 m/s, 3.3 X 105 m/s). 1) What is v, the magnitude of the velocity of the proton as it entered the region m/s Subma contalning the magnetic field? 2) What is R, the radius of curvature of the motion of the proton while it is in the region contalning the magnetikc m Submt 3) What is h, the y co ordinate of the proton as it leaves the region conating the magnetic field? m Submt 4) What is b, the z-component of the magnetic field? Note that B, is a signed number F1 F6

Explanation / Answer

Given,

D = 0.5 m ; vx = 4.1 x 10^5 m/s ; 3.3 x 10^5 m/s

1)The magnitude of v will be:

v = sqrt (vx^2 + vy^2)

v = sqrt [(4.1 x 10^5)^2 + (3.3 x 10^5)^2] = 5.26 x 10^5 m/s

Hence, v = 5.26 x 10^5 m/s

2)The radius of curvature will be:

R = D/sin(theta)

theta = tan^-1(3.3/4.1) = 38.83 deg

R = 0.5/sin38.83 = 0.8 m

Hence, R = 0.8 m

3)The h will be given by:

h = R(1 - cos(theta))

h = 0.8 (1 - cos38.83) = 0.177 m

Hence, h = 0.177 m

b)We know that magnetic force balances the centripital force

q v B = m v^2/R

B = mv/qR

B = 1.67 x 10^-27 x 5.26 x 10^5/(1.6 x 10^-19 x 0.8) = 0.00686 T

Hence, B = 6.86 x 10^-3 T

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.