A particle of mass 5.976 × 1026 kg and charge of 4.8×1019 C is accelerated from

Question

A particle of mass 5.976 × 1026 kg and charge of 4.8×1019 C is accelerated from rest in the plane of the page through a potential difference of 151 V between two parallel plates as shown. The particle is injected through a hole in the right-hand plate into a region of space containing a uniform magnetic field of magnitude 0.0827 T. The particle curves in a semicircular path and strikes a detector.

Magnetic Field B in hole Which way does the magnetic field point? 1. toward the bottom of the page 2. to the right 3. toward the lower right corner of the page 4. toward the top of the page 5. toward the upper left corner of the page 6. toward the lower left corner of the page 7. to the left 8. toward the upper right corner of the page 9. out of the page 10. into the page 012 (part 2 of 2) 10.0 points What is the magnitude of the force exerted on the charged particle as it enters the region of the magnetic field B'? Answer in units of N

Explanation / Answer

As the charge enters the magnetic field, the force is directed downwards at that very point as only then, it will move in a circle.

So,

we know that force in a magnetic field is

F=q(v*B)

So,B has to be out of the page to be the force in downward direction.

Part 2

velocity when it enters the magnetic field region will be,

0.5mv2=qV
0.5*5.976*10-26v2=4.8*10-19*151

v=49251.42446m/s

And the force will be,

F=q(v*B)

F=4.8*10-19(49251.42446*0.0827)

=1.955*10-15N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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