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konstantin123 [22]
3 years ago
6

A wire lying along a y axis form y=0 to y=0.25 m carries a current of 2.0 mA in the negative direction of the axis. the wire ful

ly lies in a magnetic field given by (0.3y)i + (0.4y)j Tesla. The magnetic force on the wire is *
Physics
1 answer:
Bess [88]3 years ago
7 0

Answer:

The force is "19 µN".

Explanation:

The lane's j-component is meaningless, as the current is flowing in the -j line.  

Therefore the power is now in the direction of + z (out of the page if x and y are in the page plane) and has the magnitude.

\ formula: \\\\\ Forec (F) = mA \\\\ \ F \ = 2.0mA \times \int {0.3} \ y \ dy  \rightarrow  \ from\ 0 \ to \ 0.25 \\\\\ F \ = \ 2.0mA \times 0.15 * 0.25^{2} m\cdot T \\\\ F = 19 \µN

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The mean may be calculated by summing the values of the refractive index and dividing the sum by the number of experiments. This is:
Mean = (1.45 + 1.56 + 1.54 + 1.44 + 1.54 + 1.53)/6
Mean = 1.51

The mean absolute error is the sum of the absolute values of errors divided by the number of trials:
MAE = (|1.45-1.51|+|1.56-1.51|+|1.54-1.51|+|1.44-1.51|+|1.54-1.51|+|1.53-1.51|)/6
MAE = 0.043

The fractional error is the MAE divided by the actual value:
Fractional error = 0.043 / 1.51
Fractional error = 43/1510

The percentage error is the fractional error multiplied by 100:
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7 0
3 years ago
A flat, circular, steel loop of radius 75 cm is at rest in a uniform magnetic field, as shown in an edge-on view in the figure (
SIZIF [17.4K]

The solution to the questions are given as

  • t=40.39 \mathrm{sec}
  • \varepsilon &=(0.12v)e^{0.057t}
  • the direction of induced current will be Counterclock vise.

<h3>What is the direction of the current induced in the loop, as viewed from above the loop.?</h3>

Given, $B(t)=(1.4 T) e^{-0.057 t}$

$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}

\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$

\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}

\varepsilon &=(0.12v)e^{0.057t}

(b) Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$

\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}

c)

In conclusion, the direction of the induced current will be Counterclockwise.

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What causes a snowflake to melt on your tongue?​
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Answer:

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When  t=2, the ball has fallen     d(2) = 16 (2²) = 64 feet .

When  t=5, the ball has fallen     d(5) = 16 (5²) = 400 feet .

Distance fallen from  t=2  until  t=5  is  (400 - 64) = 336 feet.

Time period between  t=2  until  t=5  is  (5 - 2) = 3 seconds.

Average speed of the ball from  t=2  until  t=5  is

                 (distance covered) / (time to cover the distance)

             =            336 feet        /        3 seconds       =  112 feet per second.

That's what choice-C says.        
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