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2 years ago

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Solenoid 2 has twice the diameter, twice the length, and twice as many turns as solenoid 1. How does the field B2 at the center

of solenoid 2 compare to B1 at the center of solenoid 1?

1 answer:

klemol [59]2 years ago

4 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

The correct option is option 3

Explanation:

From the question we are told that

The diameter of solenoid 1 is $d_1$

The length of solenoid 1 is $L_1$

The number of turns of solenoid is $N_1$

The diameter of solenoid 2 is $d_2 = 2d_1$

The length of solenoid 2 is $L_2 = 2L_1$

The number of turns of solenoid 2 is $N_2 = 2 N_1$

Generally the magnetic in a solenoid is mathematically represented as

$B = \frac{\mu_o * N * I }{L}$

From this equation we see that

$B \ \alpha \ \frac{N}{L}$

$B = C \frac{N}{L}$

Here C stands for constant

=> $C = \frac{B * \frac{L}{N}$

=> $\frac{B_1 * \frac{L_1}{N_1} = \frac{B_2 * \frac{L_2}{N_2}$

=> $\frac{B_1}{B_2 } = \frac{N_1 L _2}{ N_2L_1}$

=> $\frac{B_1}{B_2 } = \frac{N_1 * (2 L_1)}{ (2 N_2)L_1}$

=> $\frac{B_1}{B_2 } = 1$

=> $B_2 = B_1$

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Answer:

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Explanation:

Because only gravity is doing the work on the objects, and gravity is constant for all the objects

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1 year ago

A variable-length air column is placed just below a vibrating wire that is fixed at both ends. The length of the air column, ope

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Answer:

The speed of transverse waves in the wire is 234.26 m/s.

Explanation:

Given that,

Length of wire = 129 cm

Speed of sound in air = 340 m/s

First position of resonance = 31.2 cm

We need to calculate the wavelength

For pipe open at one end and closed at other, there is node at closed end and an anti node at open end for 1st resonance.

At 1st resonance,

$L=\dfrac{\lambda}{4}$

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Put the value into the formula

$\lambda=4\times31.2\times10^{-2}$

$\lambda=1.248\ m$

We need to calculate the frequency of sound in pipe

Using formula of frequency

$f=\dfrac{v}{\lambda}$

Put the value into the formula

$f=\dfrac{340}{1.248}$

$f=272.4\ Hz$

We need to calculate the node distance

For the wire, there are 3 segments, so 4 nodes

Node-node distance ,

$L=\dfrac{l}{3}$

$L = \dfrac{129}{3}$

$L=43\ cm$

We need to calculate the wavelength of the wire

Using formula of length

$L=\dfrac{\lambda}{2}$

$\lambda=L\times 2$

Put the value into the formula

$\lambda=43\times2$

$\lambda=86\ cm$

We need to calculate the speed of the wave

Using formula of frequency

$f=\dfrac{v}{\lambda}$

$v=f\times\lambda$

Put the value into the formula

$v=272.4\times86\times10^{-2}$

$v=234.26\ m/s$

Hence, The speed of transverse waves in the wire is 234.26 m/s.

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2 years ago

A force of 500 N is exerted on a baseball by the bat for 0.001 s. What is the change in momentum of the baseball?

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Answer: Δp = F*Δt = 500N*0.001s = 0.5Ns

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2 years ago

Two coherent sources of radio waves, A and B, are 5.00 meters apart. Each source emits waves with wavelength 6.00 meters. Consid

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Answer:

a

$z= 2.5 \ m$

b

$z = (1 \ m , 4 \ m )$

Explanation:

From the question we are told that

Their distance apart is $d = 5.00 \ m$

The wavelength of each source wave $\lambda = 6.0 \ m$

Let the distance from source A where the construct interference occurred be z

Generally the path difference for constructive interference is

$z - (d-z) = m \lambda$

Now given that we are considering just the straight line (i.e points along the line connecting the two sources ) then the order of the maxima m = 0

so

$z - (5-z) = 0$

=> $2 z - 5 = 0$

=> $z= 2.5 \ m$

Generally the path difference for destructive interference is

$|z-(d-z)| = (2m + 1)\frac{\lambda}{2}$

=> $|2z - d |= (0 + 1)\frac{\lambda}{2}$

=> $|2z - d| =\frac{\lambda}{2}$

substituting values

$|2z - 5| =\frac{6}{2}$

=> $z = \frac{5 \pm 3}{2}$

So

$z = \frac{5 + 3}{2}$

$z = 4\ m$

and

$z = \frac{ 5 -3 }{2}$

=> $z = 1 \ m$

=> $z = (1 \ m , 4 \ m )$

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2 years ago

A fisherman is fishing from a bridge and is using a "50.0-N test line." In other words, the line will sustain a maximum force of

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Answer:(a) 50 N

(b)38.34 N

Explanation:

Given

Maximum tension(T) in line 50 N

(a)If line is moving up with constant velocity i.e. there is no acceleration

This will happen when Tension is equal to weight of Fish

T-mg=0

T=mg

Maximum weight in this case will be 50 N

(b)acceleration of magnitude $2.98 m/s^2$

T-mg=ma

$50=m\left ( g+a\right )$

$50=m\left ( 12.78\right )$

m=3.91

Therefore weight is $3.91\times 9.8=38.34 N$

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