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FinnZ [79.3K]
3 years ago
15

We have a long wire with a circular cross section and radius a = 2.40 cm. The current density in this wire is uniform, with a to

tal current of 3.00 A. Find the magnitude of the magnetic field at a distance of 0.72 cm from the center axis. Treat the wire as a cylinder.
Physics
1 answer:
Alika [10]3 years ago
7 0

Answer:

The magnitude of the magnetic field is 7.49x10⁻⁶T

Explanation:

The magnetic field is:

B=(\frac{\mu i}{2\pi R^{2} } )r

Where

i = current = 3 A

R = radius = 2.4 cm = 0.024 m

r = distance = 0.72 cm = 7.2x10⁻³m

Replacing:

B=(\frac{4\pi x10^{-7} *3}{2\pi *0.024^{2} } )*7.2x10^{-3}= 7.49x10^{-6} T

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A 2.20-kg object is attached to a spring and placed on frictionless, horizontal surface. A horizontal force of 29.0 N is require
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f = 1/2π√(145 N/m/2.20 kg) = 1.29 Hz

c. maximum speed of the object

The maximum elastic potential energy of the spring = maximum kinetic  energy of the object

1/2kx² = 1/2mv²

v = (√k/m)x where v is the maximum speed of the object

v = (√145/2.2)0.2 = 1.62 m/s

d Where does the maximum speed occur?

The maximum speed occurs at  0 m

e. The maximum acceleration

a = kx/m = 145 × 0.2/2.2 = 13.2 m/s²

f. The maximum acceleration occurs at x = ± 0.2 m

g. The total energy of the system is the maximum elestic potential energy of the system

E = 1/2kx² = 1/2 × 145 × 0.2² = 2.9 J

h. When x = x₀/3

1/2k(x₀/3)² = 1/2mv²

kx₀²/9 = mv²

v = 1/3(√k/m)x₀ = 1/3(√145/2.2)0.2 = 0.54 m/s

i When x = x₀/3

a = kx₀/3m =  145 × 0.2/(2.2 × 3)= 4.39 m/s²

8 0
3 years ago
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