A charged particle leaves a potential difference with a speed of 1200m/s and is

Question

A charged particle leaves a potential difference with a speed of 1200m/s and is headed east. The particle then enters a uniform magnetic field directed into the page with magnitude 21.41T. When the particle enters the uniform field, it undergoes a specific type of motion, of which begins turning the particle toward the south.

A) If this motion has a radius of 10.68 cm and the magnitude of the charge on the particle is 3.20uC (mu), determine the mass of the particle.

B) If the path of the particle were considered a current loop, it would produce a magnetic feild of its own. Determine the magnitude and direction of the magnetic feild produced by the "current loop," at the center of the loop. What is the total magnetic field at the center of the loop, if we superpose the external feild and the field produced by the "current loop?"

Explanation / Answer

(A) Fb = m a_c

q v B = m v^2 / r

q r B = m v

(3.20 x 10^-6) (0.1068) (21.41) = (m) (1200)

m = 6.10 x 10^-9 kg ......Ans

(B) f = v / 2 pi r = 1200 / (2 x pi x0.1068)

f = 1788.3 Hz

I = q f = 5.72 x 10^-3 A  

field due to current loop,

B = u0 I / 2 R = (4pi x 10^-7)(5.72 x 10^-3) /(2 x 0.1068)

B = 3.37 x 10^-8 T ......Ans

direction = out of page

total field = (21.41) - (3.37 x 10^-8)

= 21.41 T

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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