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Aleksandr-060686 [28]
3 years ago
8

A wire carries a 11.3-mA current along the +x-axis through a magnetic field = (16.2 + 2.4 ĵ) T. If the wire experiences a force

of -15.7 N as a result, how long is the wire?
Physics
1 answer:
adelina 88 [10]3 years ago
5 0

Answer:

The length of the wire is 579 m

Explanation:

Given;

current on the wire, I =  11.3-mA

magnetic field of the wire, B = (16.2i + 2.4 ĵ) T

Magnitude of force experience by the wire, F = 15.7 N

Magnitude of force experience by  current carrying wire at a given a magnetic field strength is calculated as;

F = BILsinθ

Where;

B is magnitude of magnetic field

F is the force on the wire

L is length of the wire

θ is direction of the magnetic field

B = \sqrt{16.2^2 +2.4^2} = \sqrt{268.2} = 16.377 \ T

tan \theta = \frac{2.4}{16.2} \\\\tan \theta =  0.1482\\\\\theta = tan^{-1}(0.1482) \\\\\theta = 8.43^o

Length of the wire is calculated as;

L = \frac{F}{BIsin \theta} = \frac{15.7}{16.377*11.3*10^{-3}*sin(8.43)} = 578.9 \ m

Therefore, the length of the wire is 579 m

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The spin-drier of a washing machine revolving at 900.RPM slows down uniformly to 300.RPM while making 60. revolutions. Find the
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<h2>Answer:</h2>

<u>Acceleration is </u><u>-10.57 rad/s²  </u>

<u>Time is </u><u>6 seconds</u>

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u=900rpm= 94.248 rad/s  

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An electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric
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Answer:

1.) 11 km/s

2.) 9.03 × 10^-5 metres

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Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.

Electron q = 1.6×10^-19 C

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V = 10971.43 m/s

V = 11 Km/s approximately

(b) How far does the electron travel during the 1.3 ns interval?

The initial velocity U = 64 km/s

S = ut + 1/2at^2

S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2

S =8.32×10^-5 + 7.13×10^-6

S = 9.03 × 10^-5 metres

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