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kenny6666 [7]
3 years ago
11

Two straight wires carry current of 5A opposite direction separated by a distance of 30cm. What is the magnitude of magnetic fie

ld at point P exactly the middle of them 15cm from each wire
Physics
1 answer:
jek_recluse [69]3 years ago
6 0

Answer:

The magnitude of magnetic field of both wires is β= 13. 33x10^{-6}T

Explanation:

p=15cm*\frac{1m}{100cm}=0.15m

I=5A

u_{o} = 4*\pi x10^{-7}

Using the right hand rule field magnetic as the current go in opposite direction the field in the point exactly in the middle have the same direction both so

\beta _{t} =\beta _{1}+\beta _{2}

They have the same direction and the P point is the middle as the same current field can be find:

\beta _{t} =\beta _{1}*2

\beta = \frac{u_{o}*I }{2*\pi *p} \\\beta =\frac{4*\pi x10^{-7}*5A }{2*\pi *\frac{30}{2} } \\\beta =\frac{4x10^{-7}*5 }{0.30} \\\beta =6.6x10^{-6}T

\beta _{t} =6.6x10^{-6}*2=13.33x10^{-6}T

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6 The energy stored in the system increases.

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9  

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5 0
4 years ago
What are (a) the lowest frequency, (b) the second lowest frequency, and (c) the third lowest frequency for standing waves on a w
Delvig [45]
(a) The lowest frequency (called fundamental frequency) of a wire stretched under a tension T is given by
f_1 =  \frac{1}{2L} \sqrt{ \frac{T}{m/L} }
where
L is the wire length
T is the tension
m is the wire mass

In our problem, L=10.9 m, m=55.8 g=0.0558 kg and T=253 N, therefore the fundamental frequency of the wire is
T= \frac{1}{2L} \sqrt{ \frac{T}{m/L} }= \frac{1}{2 \cdot 10.9 m} \sqrt{ \frac{253 N}{0.0558 kg/10.9 m} }=    10.2 Hz

b) The frequency of the nth-harmonic for a standing wave in a wire is given by
f_n = n f_1
where n is the order of the harmonic and f1 is the fundamental frequency. If we use n=2, we find the second lowest frequency of the wire:
f_2 = 2 f_1 = 2 \cdot 10.2 Hz=20.4 Hz

c) Similarly, the third lowest frequency (third harmonic) is given by
f_3 = 3 f_1 = 3 \cdot 10.2 Hz = 30.6 Hz

8 0
3 years ago
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