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3 months 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:

Sav [3.1K]3 months ago

4 0

Complete Question

The entire question is displayed in the first uploaded image

Answer:

The right choice is option 3

Explanation:

The question informs us 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 $N_2 = 2 N_1$

Typically, the magnetic field in a solenoid can be expressed mathematically as

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

According to this formula we find that

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

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

Here C denotes 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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(a) Calculate the absolute pressure at the bottom of a fresh- water lake at a depth of 27.5 m. Assume the density of the water i

kicyunya [3294]

Response:

a) $P = 370.993\,kPa$ , b) $F = 25.948\,kN$

Clarification:

a) The absolute pressure at a depth of 27.5 meters is:

$P = P_{atm} + P_{man}$

$P = 101.3\,kPa + \left(1000\,\frac{kg}{m^{3}}\right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (27.5\,m)\cdot \left(\frac{1\,kPa}{1000\,Pa} \right)$

$P = 370.993\,kPa$

b) The force applied by the water is:

$F = (P - P_{atm})\cdot A$

$F = (370.993\,kPa-101.3\,kPa)\cdot \left(\frac{\pi}{4} \right)\cdot (0.35\,m)^{2}$

$F = 25.948\,kN$

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