A 3.25g bullet picks up an electric charge of 1.55 mu C as it travels down the b

Question

A 3.25g bullet picks up an electric charge of 1.55 mu C as it travels down the barrel of a rifle. It leaves the barrel at a speed of 435m/s traveling perpendicular to the earth's magnetic field, which has a magnitude of 5.50 times 10-4T. Calculate the magnitude of the magnetic force on the bullet. Express your answer with the appropriate units. Calculate the magnitude of the bullet's acceleration due to the magnetic force at the instant it leaves the rifle barrel. Express your answer with the appropriate units.

Explanation / Answer

As the velocity of the bullet is perpendicular to the direction of the magnetic field, the force on the bullet= qvB= 1.55*10^(-6)*435*5.50*10^(-4) newtons= 3.71*10^(-7) newtons Acceleration of the bullet= F/m= 3.71*10^(-7)*1000/3.25 m/s^2 =1.14*10^(-4) 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.