A negative charge of q- -7.9 times 10^-17C and m = 1.1 times 10^-26 kg enters a

Question

A negative charge of q- -7.9 times 10^-17C and m = 1.1 times 10^-26 kg enters a magnetic field B = 1.4 T with initial velocity v = 560 m/s, as shown in the figure. The magnetic field points into the screen. Randomized variables q = -7.9 times 10^-17 C m = 1.1 times 10^-26 kg B + 1.4 T v = 560 m/s Which direction will the magnetic force be on the charge? Express the magnitude of the magnetic force, F, in terms of q, v, and B. (c) Calculate the magnitude of the force F, in newtons. Under such a magnetic force, which kind of motion will the charge undergo? Express the centripetal acceleration of the particle in terms of the force F and the mass m. Calculate the magnitude of a, in meters per square second.

Explanation / Answer

Part (b)

Force=qvB

Part c

F=6.2*10(-14) N

Part d

Motion will be circular

Part e

Centripetal accleration=(v*v)/r

r=mv/qB

a=qvB/m

Part f

a=5.6*10(12) m/s(2)

Part (f)

Equalling force

mv*v/r=qvB

r=mv/qB

r=mv/qB

Part (h)

r=5.56*10(-8) m

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