A microparticle has mass of 7.5 times 10^-9 kg. The Earth\'s magnetic field at t

Question

A microparticle has mass of 7.5 times 10^-9 kg. The Earth's magnetic field at the location of microparticle is 4.9 times 10^-5 T. The charge on the microparticle is 4.0 times 10^-10 C. Compute the weight w in newtons of the microparticle. Compute v. the speed of the microparticle if its magnetic force balances its weight. Compute E, the electric field strength required for a velocity selector that allows the charged microparticle to pass through undefeated. (Assume that the magnetic field is the same as above.) If the plate separation of the velocity selector is 1.6 cm, compute the voltage across its plates. If a solenoid with 779 turns per meter produces the same magnetic field as the Earth, then compute the current through the coils.

Explanation / Answer

part A:

Weight of the microparticle = W = mg

W = 7.5*10^-9 * 9.81

W = 7.35 *10^-8 N

------------------------

part B :

use Gravitational force = magnetic force

mg = q v B

so

speed v = 7.35 *10^-8/(4*10^-10 * 4.9 *10^-5)

v = 3.75 *10^6 m/s

----------------------------------------------------------------------------

part C:

Electric field and magnetic field are related by E = Bc

so

E = 4.9 *10^-5 * 3*10^8

E = 1.47 *10^4 N/C

---------------------------

part D :

Electric field and Potential are related as V = Ed

so

V = 1.47 *10^4 * 0.016

V = 235.2 Volts

---------

part E:

magnetic field through the solonoid is

B = uo n I

Current I = 4.9 *10^-5/(4pi *10^-7 * 779)

I = 50 mA
--------------------------------

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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