A proton moving with a velocity of 4.0 x 10^4 m/s enters amagnetic field of 0.20

Question

A proton moving with a velocity of 4.0 x 10^4 m/s enters amagnetic field of 0.20 T. If the angle between the velocity ofthe proton and the direction of the magnetic field is 60 degrees,what is the magnitude of the force on the proton? A) 1.8 x 10^-15 N B) 0.60 x 10^-15 N C) 1.1 x 10^ -15 N D) 3.3 x 10^-15 N E)2.2 x 10^-15 N Why?? A proton moving with a velocity of 4.0 x 10^4 m/s enters amagnetic field of 0.20 T. If the angle between the velocity ofthe proton and the direction of the magnetic field is 60 degrees,what is the magnitude of the force on the proton? A) 1.8 x 10^-15 N B) 0.60 x 10^-15 N C) 1.1 x 10^ -15 N D) 3.3 x 10^-15 N E)2.2 x 10^-15 N Why??

Explanation / Answer

   F   =   B* q * v * sin    =   0.20 *4.0 * 104 * 1.6 * 10-19 * sin600    F   =   1.10*10-15       N    F   =   1.10*10-15       N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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