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vovikov84 [41]
3 years ago
5

If the resulting trajectory of the charged particle is a circle,what is , the angular frequency of the circular motion?Express i

n terms of ,w,m and B0 .
Physics
1 answer:
Maru [420]3 years ago
8 0

Answer:

ω = B₀q/m

Explanation:

The centripetal force on the charge equal the magnetic force on the charge. (Since the trajectory is a circle). So, mrω² = B₀qv.

v = rω The speed of the charge and r = radius of path

mrω² = B₀qrω

ω = B₀q/m

The angular frequency ω = B₀q/m

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v(t)=\frac{dx(t)}{dt} =- \omega A sin(\omega t + \phi)\\\\a(t)=\frac{dv(t)}{dt} =- \omega^2 A cos(\omega t + \phi)

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(a)

x(9.7)=1.2 cos((3 \pi *(9.7))+\frac{\pi}{5} ) \approx -0.70534m

(b)

v(9.7)=-(3\pi) (1.2) sin((3\pi *(9.7))+\frac{\pi}{5} ) \approx 9.1498 m/s

(c)

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(d)

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\phi = \frac{\pi}{5}

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f=\frac{\omega}{2 \pi} =\frac{3\pi}{2 \pi} =\frac{3}{2} =1.5 Hz

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T=\frac{2 \pi}{\omega} =\frac{2 \pi}{3 \pi} =\frac{2}{3} \approx 0.667s

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