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2 years ago

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The field B = −2ax + 3ay + 4az mT is present in free space. Find the vector force exerted on a straight wire carrying 12 A in th

e aAB direction, given A(1, 1, 1) and:
(a) B(2, 1, 1);
(b) B(3, 5, 6).

1 answer:

Nimfa-mama [501]2 years ago

5 0

The magnetic force on a current-carrying wire is given by:

F = iL×B

F = magnetic force, i = current, L = wire length vector, B = magnetic field

Note we are taking the cross product of the iL and B vectors, not the product of two scalars.

A) Given values:

i = 12A

L = AB = (2, 1, 1)m - (1, 1, 1)m = <1, 0, 0>

B = <-2, 3, 4>10⁻³T

Plug in and solve for F:

F = 12<1, 0, 0>×10⁻³<-2, 3, 4> = 10⁻³(<12, 0, 0>×<-2, 3, 4>)

Beware, we'll use array notation to show the cross product calculation:

F = 10⁻³ $\left[\begin{array}{ccc}i&j&k\\12&0&0\\-2&3&4\end{array}\right]$

F = 10⁻³<(0)(4)-(0)(3), (0)(-2)-(12)(4), (12)(3)-(0)(-2)>

F = 10⁻³<0, -48, 36>N

B) Given values:

i = 12A

L = AB = (3, 5, 6)m - (1, 1, 1)m = <2, 4, 5>

B = <-2, 3, 4>10⁻³T

Plug in and solve for F:

F = 12<2, 4, 5>×10⁻³<-2, 3, 4> = 10⁻³(<24, 48, 60>×<-2, 3, 4>)

F = 10⁻³ $\left[\begin{array}{ccc}i&j&k\\24&48&60\\-2&3&4\end{array}\right]$

F = 10⁻³<(48)(4)-(60)(3), (60)(-2)-(24)(4), (24)(3)-(48)(-2)>

F = 10⁻³<12, -216, 168>N

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laiz [17]

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here, work done = 0.
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2 years ago

A simple generator has a square armature 6.0 cm on a side. The armature has 85 turns of 0.59-mm-diameter copper wire and rotates

FrozenT [24]

Answer:

f=15.5 Hz

Explanation:

Let's determine the internal resistance:

$R=\frac{(p*L)}{A}$

ρ = 1.68*10^-8 Ω m

$L=0.060m*4*60 = 14.4m$

$A=\pi*r^2\\A= \pi*(5.9x10^-4m/2)^2=2.734*10^-7m^2$

$R=(1.68*10^-8)*(14.4m)/(2.734*10^-7m^2)= 0.884$ Ω

Since the bulb is rated at 12.0 V and 25.0 W,

Current

$I=\frac{25W}{12.0v}=2.08 A$

Therefore, voltage drop inside generator =

$V=(2.08 A)*(0.88)=2.35v$

Actual EMF required is

$E_{mf}=12.0v+2.35v=14.35v$

Note that this is an RMS value.

The peak voltage is

$v_{peak}=14.15v*\sqrt{2} =20.29v$

For a generator, by Faraday's Law,

$E_{(max)}=N*B*A*w$

$20.29v=(60)*(0.650T)*(0.06m)^2$ *ω

ω $=144.5\frac{rad}{s}$

f=ω/(2π)=

f=144.5 rad/s/(2π)

f=23.001 Hz

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2 years ago

The absolute pressure, in kilopascals, a depth 10m below sea level is most nearly?

saul85 [17]

Answer:

option A

Explanation:

given,

depth of the sea level = 10 m

g = 10 m/s²

Pressure underwater = ?

we know,

P = ρ g h

where ρ is the density of water which is equal to 1000 kg/m³

h is the depth of sea level

P = ρ g h

P = 1000 x 10 x 10

P = 100000 Pa

P = 100 kPa

Hence, the correct answer is option A

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2 years ago

Maximum voltage produced in an AC generator completing 60 cycles in 30 sec is 250V. (a) What is period of armature? (b) How many

Vlada [557]

Answer:

a. 2 Hz b. 0.5 cycles c . 0 V

Explanation:

a. What is period of armature?

Since it takes the armature 30 seconds to complete 60 cycles, and frequency f = number of cycles/ time = 60 cycles/ 30 s = 2 cycles/ s = 2 Hz

b. How many cycles are completed in T/2 sec?

The period, T = 1/f = 1/2 Hz = 0.5 s.

So, it takes 0.5 s to complete 1 cycles. At t = T/2 = 0.5/2 = 0.25 s,

Since it takes 0.5 s to complete 1 cycle, then the number of cycles it completes in 0.25 s is 0.25/0.5 = 0.5 cycles.

c. What is the maximum emf produced when the armature completes 180° rotation?

Since the emf E = E₀sinθ and when θ = 180°, sinθ = sin180° = 0

E = E₀ × 0 = 0

E = 0

So, at 180° rotation, the maximum emf produced is 0 V.

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2 years ago

Emmy kicks a soccer ball up at an angle of 45° over a level field. She watches the ball's trajectory and notices that it lands,

Elenna [48]

Let $u$ be the initial velocity of the soccer ball at an angle of inclination of $\theta_0$ with the positive x-axis.

Given that:

$\theta_0=45^{\circ}$

The horizontal distance covered by the projectile=20 m

Time of flight, $t_f=2$ seconds

Acceleration due to gravity, $g= 10 m/s^2$ downward.

As "north" and "up" as the positive x ‑ and y ‑directions, respectively.

So, $g= -10 m/s^2$

As the acceleration due to gravity is in the vertical direction, so the horizontal component of the initial velocity remains unchanged.

The x-component of the initial velocity, $u_x=u\cos\theta_0$ .

The horizontal distance covered by the projectile $= u_x\times t_f$

$\Rightarrow u_x\times t_f=20$

$\Rightarrow u_x\times 2=20$

$\Rightarrow u_x=10$ m/s

So, the horizontal component of the velocity is 10 m/s which is constant and the graph has been shown in the figure (i).

Now, $u\cos(45^{\circ})=10$ [as $u_x=u\cos\theta_0$ ]

$\Rightarrow u=10\sqrt{2}$ m/s.

The vertical component of the initial velocity,

$u_y= u\sin\theta_0$

$\Rightarrow u_y=10\sqrt{2}\sin(45^{\circ})$

$\Rightarrow u_y=10$ m/s

Let v be the vertical component of the velocity at any time instant t.

From the equation of motion,

$v=u+at$

where u: initial velocity, v: final velocity, a: constant acceleration, and t: time taken to change the velocity from u to v.

In this case, we have $u=u_y, a= -10 m/s^2$ .

So at any time instant, t.

$v=u_y+(-10)t$

$\Rightarrow v=10-10t$

The vertical component of the velocity, v, is the function of time and related as $v=10-10t$ .

This is a linear equation.

At 2 second, the vertical component of the velocity

v=10-10x2=-10 m/s.

The graph has been shown in figure (ii).

7 0

2 years ago