Ask question

Ask question

All categories

3 years ago

10

Marvin uses a long copper wire with resistivity 1.68 x 10^-8 Ω⋅m and diameter 1.00 x 10^−3 m to create a solenoid that has a 3

.00 cm radius.
Part A: Calculate the total resistance if Marvin uses 11.3 meters of wire to create the solenoid.

Part B: Calculate how many turns are in the solenoid.

Part C: A 3.00 Volt potential difference is applied across the wires of the solenoid. Calculate the current that passes through the solenoid.

Part D: If the coils in the solenoid are separated so that they are evenly spaced across a total length of 0.1 m, what is the magnitude of the magnetic field along the central axis inside of the solenoid?

2 answers:

77julia77 [94]3 years ago

7 0

Answer:

Explanation:

resistivity of copper, ρ = 1.68 x 10^-8 ohm - m

diameter = 1 x 10^-3 m

radius of wire, r = 0.5 x 10^-3 m

Radius of solenoid, r' = 3 cm

(A)

Length of wire, l = 11.3 m

Let the resistance of wire is R.

$R= \frac{\rho\times l}{A}$

where, A is the crossection of wire

A = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3 = 7.85 x 10^-7 m²

$R= \frac{1.68\times 10^{-8}\times 11.3}{7.85\times 10^{-7}}$

R = 0.24 ohm

(B)

Let the number of turns is N.

Length of wire = N x circumference of one turn

11.3 = N x 2 x 3.14 x 0.03

N = 60

(C)

V = 3 V

R = 0.24 ohm

V = i x R

3 = i x 0.24

i = 12.5 A

(D) number of turns per unit length, n = N / 0.1 = 60 / 0.1 = 600

The magnetic field is given by

$B = \mu _{0}ni$

$B = 4 \times 3.14 \times 10^{-7}\times 600\times 12.5$

B = 9.42 x 10^-3 Tesla

Mazyrski [523]3 years ago

4 0

Answer with Explanation:

We are given that

Resistivity of copper wire= $\rh0=1.68\times 10^{-8}\Omega m$

Diameter=d= $1.00\times 10^{-3} m$

Radius of copper wire= $r=\frac{d}{2}=\frac{1}{2}\times 10^{-3} m$

Radius of solenoid=r' $=3 cm=3\times 10^{-2} m$

1 m=100 cm

a.Length of wire=l=11.3 m

Area of wire=A= $\pi r^2$

Where $\pi=3.14$

A= $3.14\times (\frac{1}{2}\times 10^{-3})^2$

Resistance, R= $\rho \frac{l}{A}$

Using the formula

$R=1.68\times 10^{-8}\times\frac{11.3}{3.14\times (\frac{1}{2}\times 10^{-3})^2}$

$R=0.24\Omega$

B.Length of solenoid= $2\pi r'=2\times 3.14\times 3\times 10^{-2}=0.188$ m

Number of turns= $n_0=\frac{l}{2\pi r'}=\frac{11.3}{0.188}$

$n_0$ =60

C.Potential difference,V=3 V

Current,I= $\frac{V}{R}$

I= $\frac{3}{0.24}=12.5 A$

D.Total length =0.1 m

Number of turns per unit length,n= $\frac{60}{0.1}=600$

Magnetic field along central axis inside of the solenoid,B= $\mu_0 nI$

$B=4\pi\times 10^{-7}\times 12.5\times 600=9.42\times 10^{-3} T$

You might be interested in

A wheel 1.0 m in radius rotates with an angular acceleration of 4.0rad/s2 . (a) If the wheel’s initial angular velocity is 2.0 r

Oliga [24]

Answer:

(a) ωf= 42 rad/s

(b) θ = 220 rad

(c) at = 4 m/s² , v = 42 m/s

Explanation:

The uniformly accelerated circular movement, is a circular path movement in which the angular acceleration is constant.

There is tangential acceleration (at ) and is constant.

We apply the equations of circular motion uniformly accelerated :

ωf= ω₀ + α*t Formula (1)

θ= ω₀*t + (1/2)*α*t² Formula (2)

at = α*R Formula (3)

v= ω*R Formula (4)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular velocity ( rad/s)

ωf: final angular velocity ( rad/s)

R : radius of the circular path (cm)

at : tangential acceleration (m/s²)

v : tangential speed (m/s)

Data

α = 4.0 rad/s² : wheel’s angular acceleration

t = 10 s

ω₀ = 2.0 rad/s : wheel’s initial angular velocity

R = 1.0 m : wheel’s radium

(a) Wheel’s angular velocity after 10 s

We replace data in the formula (1):

ωf= ω₀ + α*t

ωf= 2 + (4)*(10)

ωf= 42 rad/s

(b) Angle that rotates the wheel in the 10 s interval

We replace data in the formula (2):

θ= ω₀*t + (1/2)*α*t²

θ= (2)*(10) + (1/2)*(4)*(10)²

θ= 220 rad

θ= 220 rad

(c) Tangential speed and acceleration of a point on the rim of the wheel at the end of the 10-s interval

We replace data in the Formula (3)

at = α*R = (4)(1)

at = 4 m/s²

We replace data in the Formula (4)

v= ω*R = (42)*(1)

v = 42 m/s

6 0

3 years ago

Which of the following choices concerning the net magnetic flux through any enclosed surface is true according to Gauss' law for

11Alexandr11 [23.1K]

Answer:

e. The net magnetic flux in this case would be equal to zero.

Explanation:

As per Gauss law of magnetism we need to find the net magnetic flux through a closed loop

here we know that net magnetic flux is the scalar product of magnetic field vector and area vector

so here we have

$\int B. dA$ = net magnetic flux

since we know that magnetic field always forms closed loop so if we find the integral over a closed loop

then in that case the value of the close integral must be zero

so correct answer would be

e. The net magnetic flux in this case would be equal to zero.

6 0

3 years ago

A student places two books on a table. One book weighs less than the other book

Gnoma [55]

The book that weighs less

6 0

2 years ago

The picture below shows a football player kicking a football. This is known as two dimensional motion. In which direction does t

Molodets [167]

You are true, Stokey. The correct choice is<em> (c)</em>.

If you think about it a little bit more, you'll realize that "side to side" is also a horizontal motion.

Anyway, the question told us that we're only talking about 2 dimensions. Well, the The measurements in the 2 dimensions are the x and the y . If there were a z measurement, that would be the 3rd dimension.

7 0

3 years ago

In Anchorage, collisions of a vehicle with a moose are so common that they are referred to with the abbreviationMVC. Suppose a 1

lara31 [8.8K]

Answer:

Part a)

$f = \frac{8}{9}$

Part b)

$f = \frac{120}{169}$

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

Explanation:

Part a)

Let say the collision between Moose and the car is elastic collision

So here we can use momentum conservation

$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$1000 v_o = 1000 v_{1f} + 500 v_{2f}$

also by elastic collision condition we know that

$v_{2f} - v_{1f} = v_o$

now we have

$2v_o = 2v_{1f} + v_o + v_{1f}$

now we have

$v_{1f} = \frac{v_o}{3}$

Now loss in kinetic energy of the car is given as

$\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)$

$\Delta K = \frac{1}{2}m(v_o^2 - \frac{v_o^2}{9})$

so fractional loss in energy is given as

$f = \frac{\Delta K}{K}$

$f = \frac{\frac{4}{9}mv_o^2}{\frac{1}{2}mv_o^2}$

$f = \frac{8}{9}$

Part b)

Let say the collision between Camel and the car is elastic collision

So here we can use momentum conservation

$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$1000 v_o = 1000 v_{1f} + 300 v_{2f}$

also by elastic collision condition we know that

$v_{2f} - v_{1f} = v_o$

now we have

$10v_o = 10v_{1f} + 3(v_o + v_{1f})$

now we have

$v_{1f} = \frac{7v_o}{13}$

Now loss in kinetic energy of the car is given as

$\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)$

$\Delta K = \frac{1}{2}m(v_o^2 - \frac{49v_o^2}{169})$

so fractional loss in energy is given as

$f = \frac{\Delta K}{K}$

$f = \frac{\frac{60}{169}mv_o^2}{\frac{1}{2}mv_o^2}$

$f = \frac{120}{169}$

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

8 0

3 years ago