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4 years ago

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Two coils that are separated by a distance equal to their radius and that carry equal currents such that their axial fields add

are called Helmholtz coils. A feature of Helmholtz coils is that the resultant magnetic field between the coils is very uniform. Let R = 11.0 cm, I = 17.0 A, and N = 300 turns for each coil. Place one coil in the y-z plane with its center at the origin and the other in a parallel plane at R = 11.0 cm. Calculate the resultant field Bx at x1 = 2.8 cm, x2 = 5.5 cm, x3 = 7.3 cm, and x4= 11.0 cm.

1 answer:

jok3333 [9.3K]4 years ago

4 0

Answer:

When x = 2.8 cm, $B_{x1} = 0.0265 T$

When x = 5.5 cm, $B_{x2} = 0.0209 T$

when x = 7.3 cm, $B_{x3} = 0.0169 T$

When x = 11.0 cm, $B_{x4} = 0.0103 T$

Explanation:

According to Biot-Savart law,

$B_{x} = \frac{N \mu_{o}IR^{2} }{2(x^{2} +R^{2} )^{3/2} }\\$ .......................(1)

R = 11.0 cm = 0.11 m

I = 17.0 A

N = 300 turns

$\mu_{o} = 4\pi * 10^{-7} N/A^{2}$

When x₁ = 2.8 cm = 0.028 m

$B_{x1} = \frac{300 *(4\pi * 10^{-7} ) * 17 *0.11^{2} }{2(0.028^{2} +0.11^{2} )^{3/2} }\\B_{x1} = 0.0265 T$

When x₂ = 5.5cm = 0.055 m

$B_{x2} = \frac{300 *(4\pi * 10^{-7} ) * 17 *0.11^{2} }{2(0.055^{2} +0.11^{2} )^{3/2} }\\B_{x2} = 0.0209 T$

When x₃ = 7.3 cm = 0.073 m

$B_{x3} = \frac{300 *(4\pi * 10^{-7} ) * 17 *0.11^{2} }{2(0.073^{2} +0.11^{2} )^{3/2} }\\B_{x3} = 0.0169 T$

When X₄ = 11.0 cm = 0.11 m

$B_{x4} = \frac{300 *(4\pi * 10^{-7} ) * 17 *0.11^{2} }{2(0.11^{2} +0.11^{2} )^{3/2} }\\B_{x4} = 0.0103 T$

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Mademuasel [1]

Answer:

39.6138 °C

Explanation:

Heat gain by water = Heat lost by nickel

Thus,

$m_{water}\times C_{water}\times (T_f-T_i)=-m_{nickel}\times C_{nickel}\times (T_f-T_i)$

Where, negative sign signifies heat loss

Or,

$m_{water}\times C_{water}\times (T_f-T_i)=m_{nickel}\times C_{nickel}\times (T_i-T_f)$

For water:

Volume = 100.0 mL

Density of water= 1 g/mL

So, mass of the water:

$Mass\ of\ water=Density \times {Volume\ of\ water}$

$Mass\ of\ water=1 g/mL \times {100.0\ mL}$

Mass of water = 100 g

Initial temperature = 22.0 °C

Specific heat of water = 4.186 J/g°C

For nickel:

Mass = 275 g

Initial temperature = 100 °C

Specific heat of nickel = 0.444 J/gK = 0.444 J/g°C

So,

$100\times 4.186\times (T_f-22.0)=275\times 0.444\times (100-T_f)$

$418.6\times (T_f-22.0)=122.1\times (100-T_f)$

$418.6\times T_f+122.1\times T_f=21419.2$

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The final temperature of the combined metals is 39.6138 °C

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3 years ago

Read 2 more answers

A sample of gas has a volume of 25L at a pressure of 200 kPa and a temperature of 25°C. What would be the volume if the pressure

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Answer:

18 L

Explanation:

Use ideal gas law:

PV = nRT

Find initial number of moles, but first convert to units that match the gas constant units.

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4 years ago

What is 34 + (5) × (1.2465) written with the correct number of significant figures?

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Answer: 40

Explanation:

= 34 + 5 * 1.2465

= ‭40.2325‬

= 40

The number of significant figures in the answer should be the same as the number with the least number of significant figures that any of the digits in the equation have.

32 has 2 significant figures so the answer has to be 2 significant figures which is 40.

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3 years ago

a proton of mass 1 u travelling with a speed of 3.6 x 10 ^4 m/s has an elastic head on collision with a helium nucleus initially

CaHeK987 [17]

Answer:

Velocity of the helium nuleus = 1.44x10⁴m/s

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Explanation:

From the conservation of linear momentum of the proton collision with the He nucleus:

$P_{1i} + P_{2i} = P_{1f} + P_{2f]$ (1)

<em>where $P_{1i}$ : is the proton linear momentum initial, $P_{2i}$ : is the helium nucleus linear momentum initial, $P_{1f}$ : is the proton linear momentum final, $P_{2f}$ : is the helium nucleus linear momentum final </em>

<u>From (1):</u>

$m_{1}v_{1i} + 0 = m_{1}v_{1f} + m_{2}v_{2f}$ (2)

<em>where m₁ and m₂: are the proton and helium mass, respectively, $v_{1i}$ and $v_{2i}$ : are the proton and helium nucleus velocities, respectively, before the collision, and $v_{1f}$ and $v_{2f}$ : are the proton and helium nucleus velocities, respectively, after the collision </em>

By conservation of energy, we have:

$K_{1i} + K_{2i} = K_{1f} + K_{2f}$ (3)

<em>where $K_{1i}$ and $K_{2i}$ : are the kinetic energy for the proton and helium, respectively, before the colission, and $K_{1f}$ and $K_{2f}$ : are the kinetic energy for the proton and helium, respectively, after the colission </em>

<u>From (3):</u>

$\frac{1}{2}m_{1}v_{1i}^{2} + 0 = \frac{1}{2}m_{1}v_{1f}^{2} + \frac{1}{2}m_{2}v_{2f}^{2}$ (4)

<u>Now we have two equations: (2) ad (4), and two incognits: $v_{1f}$ and $v_{2f}$ . </u>

Solving equation (2) for $v_{1f}$ , we have:

$v_{1f} = v_{1i} -\frac{m_{2}}{m_{1}} v_{2f}$ (5)

<u>From getting (5) into (4) we can obtain the $v_{2f}$ :</u>

$v_{2f}^{2} \cdot (\frac{m_{2}^{2}}{m_{1}} + m_{2}) - 2v_{2f}v_{1i}m_{2} = 0$

$v_{2f}^{2} \cdot (\frac{(4u)^{2}}{1u} + 4u) - v_{2f}\cdot 2 \cdot 3.6 \cdot 10^{4} \cdot 4u = 0$

From solving the quadratic equation, we can calculate the velocity of the helium nucleus after the collision:

$v_{2f} = 1.44 \cdot 10^{4} \frac{m}{s}$ (6)

Now, by introducing (6) into (5) we get the proton velocity after the collision:

$v_{1f} = 3.6 \cdot 10^{4} -\frac{4u}{1u} \cdot 1.44 \cdot 10^{4}$

$v_{1f} = -2.16 \cdot 10^{4} \frac{m}{s}$

The negative sign means that the proton is moving in the opposite direction after the collision.

I hope it helps you!

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