Question

A proton orbits a long charged wire, making 1.80 ×106 revolutions per second. The radius of the orbit is 1.20 cm What is the wire's linear charge density?.

Answer 1

Answer:

linear charge density = -9.495 × $10^{-34}$ C/m

Explanation:

given data

revolutions per second = 1.80 × $10^{6}$

radius = 1.20 cm

solution

we know that when proton to revolve around charge wire then centripetal force is require to be in orbit of radius around provide by electric force

so

- q × E = m × w² × r     ..................1

- 9 × $10^{9}$  × $\frac{2*linear\ charge\ density}r}$ q =  m × w² × r   ............2

and w = $\frac{2*\pi}{T}$  

w = $\frac{d\theta }{dt}$

w = 1.80 × $10^{6}$ × $\frac{2*\pi}{1}$

w = 11304000 rad/s

so here from equation 2

- 9 × $10^{9}$  × $\frac{2*linear\ charge\ density}{0.012}$ 1.80 × $10^{6}$ =  1.672 × $10^{-27}$ × 11304000² × 0.0120  

linear charge density = -9.495 × $10^{-34}$ C/m

Answer 2

The linear charge density is $2.14*10^{-18} C/m$

The linear charge density is given as,

                     Linear charge density$=\frac{Q}{L}$

Where Q is total charge and L is length.

Charge on one proton $Q=1.6*10^{-19} C$

L = circumference of orbit

$L=2\pi r=2*3.14*1.2*10^{-2}=0.075m$

Substitute value in above equation.

Linear charge density is,

                      $\frac{Q}{L}=\frac{1.6*10^{-19} }{0.075} =2.14*10^{-18} C/m$

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