Step gently onto a bathroom scale, read the dial, and then jump from a chair onto the same scale.

If the wire is replaced by an infinite current sheet with density Js = 0.40 A/m, what would be the magnetic flux (in T · m2) thr

Sav [3153]

Answer:

Clarification: $\phi _{B} =0.855 T-m^{-2}$

Considering the following data the density of the current sheet is 0.40 A/m

The length a = 0.27 m

and the width b = 0.63 m

For an infinite sheet, the magnetic field is described as

The magnetic flux is given as

$B = \mu _{O}J$

magnetic flux =

$\phi _{B} = BA$

$= \mu _{O}Jab$

$= 4\pi *0.40*0.27*0.63$

$\phi _{B} =0.855 T-m^{-2}$

6 0

2 months ago

Wire A has the same length and twice the radius of wire B. Both wires are made of the same material and carry the same current.

serg [3582]

Answer:

$V_A= \frac{I_A}{n_A e A_A}= \frac{I}{ne 4A_B}= \frac{1}{4} \frac{I}{neA_B}$

$V_B= \frac{I_B}{n_B e A_B}= \frac{I}{ne A_B}$

Consequently, we find that:

$V_A = \frac{1}{4} V_B$

Thus, the most suitable answer would be:

a. vA = vB/4

Explanation:

In this situation, we can establish the following conditions:

$L_A = L_B =L$ both wires share the same length

both wires carry an identical current $I_A = I_B =I$

Both wires are constructed of the same material, indicating that the electron density (n) remains constant across both wires

$n_A = n_B =n$

We also know that $r_A = 2 r_B$ where r signifies the radius.

Given that wires are cylindrical in shape, we can determine the area for each case:

$A_A= \pi r^2_A = \pi (2r_B)^2 = 4 \pi r^2_B= 4 A_B$

$A_B = \pi r^2_B$

Thus, we conclude that

$A_A = 4 A_B$

Now we are aware that the drift velocity of an electron in a wire can be described by:

$v_d = \frac{I}{neA}$

Where I denotes the current, n is the electron density, e represents the electron charge, and A signifies the area.

By substituting, we arrive at:

$V_A= \frac{I_A}{n_A e A_A}= \frac{I}{ne 4A_B}= \frac{1}{4} \frac{I}{neA_B}$

$V_B= \frac{I_B}{n_B e A_B}= \frac{I}{ne A_B}$

So we observe that:

$V_A = \frac{1}{4} V_B$

Thus, the most fitting answer is:

a. vA = vB/4

6 0

2 months ago

According to a rule-of-thumb. every five seconds between a lightning flash and the following thunder gives the distance to the f

ValentinkaMS [3465]

Answer:

$S_{s}=300 m/s$

According to the guideline for kilometers, every three seconds between a lightning strike and the subsequent thunder indicates the distance to the flash in kilometers.

Explanation:

To calculate the speed of sound in meters per second, we need to utilize certain conversion factors. One mile corresponds to 5 seconds after witnessing the lightning. Furthermore, 1 mile comprises 5280 feet, and 1 foot is equivalent to 0.3048 meters. This information is sufficient to solve the issue. The conversion ratios can be set up like this:

$\frac{1mi}{5s}*\frac{5280ft}{1mi}*\frac{0.3048m}{1ft}=321.87m/s$

Observe how the ratios are organized such that the units cancel out during calculations. One ratio has miles in the numerator while the other has them in the denominator, leading to cancellation. The same applies to the feet.

The question requires us to provide the answer to one significant figure, resulting in the speed of sound rounding to 300m/s.

For the second part, we will again utilize conversions. This time we will set our ratios in reverse and realize that there are 1000 meters in 1 kilometer, leading us to:

$\frac{5s}{1mi}*\frac{1mi}{5280ft}*\frac{1ft}{0.3048m}*\frac{1000m}{1km}=3.11s/km$

This signifies that for every 3.11 seconds, the distance to the lightning strike is 1 kilometer. Since this is a fabric of general knowledge, we round to the nearest whole number for simplicity, establishing the guideline:

According to the rule for kilometers, every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.

3 0

3 months ago

The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is sudde

Maru [3345]

Explanation:

Data provided:

Area A = 10 cm×2 cm = 20×10⁻⁴ m²

Separation distance d between the plates = 1 mm = 1×10⁻³ m

Battery voltage, or emf = 100 V

Resistance = 1025 ohm

Solution:

In an RC circuit, the voltage across the plates varies with time t. At the outset, the voltage matches that of the battery, V₀ = emf = 100V. However, after a certain time t, both the resistance and capacitance alter this, leading to a final voltage V expressed as

$V = V_{0}(1-e^{\frac{-t}{RC} } )\\\frac{V}{V_{0} } = 1-e(^{\frac{-t}{RC} }) \\e^{\frac{-t}{RC} } = 1- \frac{V}{V_{0} }$