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larisa86 [58]
3 years ago
12

A superconducting solenoid has 3300 turns per meter and can carry a maximum current of 4.1 ka. find the magnetic field strength

in the solenoid.

Physics
2 answers:
jeyben [28]3 years ago
8 0
B = μ x n x I = 1.25 x 10^-6 x 3300 x 4.1 x 10^3
                    =  16.91 T
avanturin [10]3 years ago
7 0
Hope this helps you.

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Read 2 more answers
A parallel-plate capacitor has 2.10 cm × 2.10 cm electrodes with surface charge densities ±1.00×10-6 C/m2. A proton traveling pa
Darya [45]

Answer:

x=0.53x10^{-3} m

Explanation:

Using Gauss law the field is uniform so

E=ζ/ε

Charge densities ⇒ζ=1.x10x^{-6} \frac{C}{m^{2}}

ε=8.85x10^{-12} \frac{C^{2}}{n*m^{2}}

E=\frac{1x10^{-6}\frac{C}{m^{2}}}{8.85x^{-12}\frac{C^{2} }{N*m^{2}}} \\E=0.11299 x10^{-6} \frac{N}{C}

Force of charge is

F_{q}=q*E\\F_{q}=1.6x10^{-19}C*0.11299x10^{6}\frac{N}{C} \\F_{q}=1.807x10^{-14} N

F_{q}=m*a\\a=\frac{F_{q}}{m}=\frac{1.807x10^{-13}N}{1.67x10^{-27}}\\ a=1.082x0^{14} \frac{m}{s^{2}} \\t=\frac{x}{v}\\ x=2.1cm\frac{1m}{100cm}=0.021m \\v=6.7x10^{6}\frac{m}{s} \\ t=\frac{0.021m}{6.7x10^{6}\frac{m}{s}} \\t=3.13x10^{-9}s

So finally knowing the acceleration and the time the distance can be find using equation of uniform motion

x_{f}=x_{o}+\frac{1}{2}*a*t^{2}\\ x_{o}=0\\x_{f}=\frac{1}{2} a*t^{2}=\frac{1}{2}*1.082x10^{14}\frac{m}{s^{2} } *(3.134x^{-9}s)^{2}  \\x_{f}=0.53x^{-3}m

5 0
3 years ago
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