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Taya2010 [7]
3 years ago
14

A 2.4 mm -diameter copper wire carries a 37 A current (uniform across its cross section). Determine the magnetic field at the su

rface of the wire. Express your answer using two significant figures
Physics
1 answer:
cluponka [151]3 years ago
7 0

Answer:

Explanation:

We shall apply Ampere's circuital law to find out magnetic field . It is given as follows.

∫B.dl = μ₀ I , B is magnetic field , I is current ,  μ₀ is permeability .

Radius of the wire r = 1.2 x 10⁻³ m

magnetic field B will be circular in shape around the wire. If B is uniform

∫B.dl = B x 2πr  

B x 2πr  = μ₀ I

B = μ₀ I / 2πr

= 4π x 10⁻⁷ x 37 /2πx1.2 x 10⁻³

= 10⁻⁷ x 2x37 / 1.2 x 10⁻³

= 61.67 x 10⁻⁴ T

= 62  x 10⁻⁴ T

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An electron in an electron gun is accelerated from rest by a potential of 25 kV applied over a distance of 1 cm.
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Answer:

9.38\times 10^7 m/s

Explanation:

We are given that

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We have to find the final velocity of the electron.

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shutvik [7]

Answer:

a = 6.53 m/s^2

v = 11.5689 m/s

Explanation:

Given data:

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P =f_{max} v

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v =0.7 \times \frac{P_{max}}{ma_{max}}

v = 0.7 \frac{217 [\frac{746 w}{1 hp}]}{1500 \times 6.53}

v = 11.5689 m/s

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