A ionized atom moves out of a split at point S into a magnetic field B. The atom

Question

A ionized atom moves out of a split at point S into a magnetic field B. The atom has a speed of v = 3.00 * 10^6 m/s The nucleus of the atom contains one proton and has a mass of m = 1.67 10^-27 kg. The charge on the atom is e. The hemisphere that the atom maps out after entering the magnetic field is illustrated below, . If the radius of curvature of the hemisphere is R = 0.300 m: What is the value of the magnetic field B? W hat is the Value of the magnetic force on the charge. What is the Value of the centripetal acceleration of the charge as it moves through the fieId An ionized atom leaves point S and travels in path of hemisphere to point P.

Explanation / Answer

here,

v = 3 * 10^6 m/s

m = 1.67 * 10^-27 kg

R = 0.3 m

a)

let the magnetic feild be B

R = m * v /( q * B)

0.3 = 1.67 * 10^-27 * 3 * 10^6 /( 1.6 * 10^-19 * B)

B = 1.04 * 10^-1 T

b)

the magnetic force , F = q * v * B

F = 1.6 * 10^-19 * 3 * 10^6 * 0.104

F = 4.99 * 10^-20 N

c)

the centripital accelration , a = F/m

a = 4.99 * 10^-20 /( 1.67 * 10^-27)

a = 3 * 10^7 m/s^2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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