500 km/s experiences a magneticforce of 0.00320 nN vertically downward. Find the

Question

500 km/s experiences a magneticforce of 0.00320 nN vertically downward. Find themagnitude of the weakest magnetic field required to produce thisforce.

PART B
An electron moves in a uniform, horizontal 2.10-T magnetic field that istoward the west. What must the magnitude of the minimum velocity ofthe electron be so that magnetic force on it will be 4.60pN, vertically upward? What is the direction of the minimalvelocity?


2.10-T magnetic field that istoward the west. What must the magnitude of the minimum velocity ofthe electron be so that magnetic force on it will be 4.60pN, vertically upward? What is the direction of the minimalvelocity?

Explanation / Answer

Charge q = 8 e = 8 * 1.6 * 10 ^ -19 C speed v = 500* 10 ^ 3 m / s force F = 0.0032* 10 ^ -9 N we know F = Bvq from this magnetic field B = F / vq                                       = 5 T (B). magnetic field B = 2.1 T force F = 4.6 * 10 ^ -12 N charge of electron q = 1.6 * 10 ^ -19 C we know F = Bvq from this minimum velocity v = F / Bq                                             =1.369 * 10 ^ 7 m / s

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