Question

A duck flying horizontally due north at 12.3 m/s passes over East Lansing, where the vertical component of the Earth's magnetic field is 4.78×10-5 T (pointing down, towards the Earth). The duck has a positive charge of 7.64×10-8 C. What is the magnitude of the magnetic force acting on the duck?

Answer 1

Answer:

4.49 x 10^-11 newton

Explanation:

v = 12.3 m/s along north = 12.3 j m/s

B = 4.78 x 10^-5 T downwards = 4.78 x 10^-5 k T

q = 7.64 x 10^-8 C

force on a charged particle when it is moving in a uniform magnetic field is given by

F = q (v x B )

F = 7.64 x 10^-8 {(12.3 i) x (4.78 x 10^-5 k)}

F = 4.49 x 10^-11 (- k) newton

magnitude of force = 4.49 x 10^-11 newton