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gregori [183]
3 years ago
11

The magnitude of the magnetic field that a long and extremely thin current-carrying wire produces at a distance of 3.0 µm from t

he center of the wire is 2.0 × 10-3 T. How much current is flowing through the wire?
Physics
2 answers:
crimeas [40]3 years ago
7 0

Answer: 0.03002A

Explanation: The formulae that relates the magnetic field strength B at a point (r) away from the center of a conductor carrying a current of value (I) is given below as

B = Uo×I/2πr

From our question, B =2.0×10^-3 T, r = 3.0×10^-6m

I =?, Uo = permeability of free space = 1.256×10^-6 mkg/s²A².

By substituting the parameters, we have that

2×10^-3 = 1.256×10^-6 × I/2π(3.0×10^-6)

2×10^-3 × 2π(3.0×10^-6) = 1.256×10^-6 × I

3.77×10^-8 = 1.256×10^-6 × I

I = 3.77×10^-8/ 1.256×10^-6

I = 3.002×10^-2 = 0.03002A

Reil [10]3 years ago
6 0

Answer:

Current (I) = 3 x 10^-2 A

Explanation:

As we know, B = 4\pi 10^-7 *l/ 2\pi r

By putting up the values needed from the data...

Current (I) = 2 x 3.14 x (3.0 x 10^-6) (2.0 x 10^-3) / 4 x 3.14 x 10^-7 = 3 x 10^-2 A

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Technician A says that in a series circuit, if one light goes out, the rest of the lights stay on. Technician B says that in a p
vivado [14]

Answer:

Explanation:

Technician A is incorrect as in series circuit the current is the same for all the resistors and if one bulb fails in the path then the circuit is incomplete and therefore no flow of current.

Technician B is correct as in parallel circuit there are multiple paths for current through which it can pass and malfunctioning of one resistor does not affect the other.      

6 0
3 years ago
A metal wire has a resistance of 14.00 Ω at a temperature of 25.0°C. If the same wire has a resistance of 14.55 Ω at 90.0°C, wha
aliina [53]

Answer:

13.52 Ω

Explanation:

coefficient of thermal resistance be α

R₀ , R₂₅ , R₉₀ and R₋₃₂ be resistances at 0 , 25 , 90 , and - 32 degree

R₂₅ = R₀ + α x 25

R₉₀ = R₀ + α x 90

R₉₀ - R₂₅ = 65 x α

α = (R₉₀ - R₂₅ )/ 65

= (14.55 - 14) / 65

=   .55 / 65 Ω per °C,

R₂₅ = R₀ + α x 25

14 = R₀ + (.55 / 65 )x 25

=  R₀ + .2115

R₀ = 13.7885 Ω

R₋₃₂ = R₀ - α x 32

= 13.7885 -(  .55 / 65) x 32

=  13.7885 - .27077

= 13.51773 Ω

= 13.52 Ω

7 0
3 years ago
Read 2 more answers
2-Pema runs
Oxana [17]

Answer:

20 m/s

Explanation:

Distance covered by man =500m

Time taken by man=25 seconds

Speed of man = distance/time

= 500/25

=20 m/s

4 0
3 years ago
The density of a certain type of plastic is 0.75 g/cm3 . If a sheet of this plastic is 10.0 m long, 1.0 m wide, and 1 cm thick,
Oksanka [162]

Answer:

7.5 g

Explanation:

8 0
3 years ago
A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i
Arturiano [62]

Answer:

q / q_{1} = 2, assuming that q_{1} and (q - q_{1}) are point charges.

Explanation:

Let k denote the coulomb constant. Let r denote the distance between the two point charges. In this question, neither k and r depend on the value of q_{1}.

By Coulomb's Law, the magnitude of electrostatic force between q_{1} and (q - q_{1}) would be:

\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}.

Find the first and second derivative of F with respect to q_{1}. (Note that 0 < q_{1} < q.)

First derivative:

\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}.

Second derivative:

\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}.

The value of the coulomb constant k is greater than 0. Thus, the value of the second derivative of F with respect to q_{1} would be negative for all real r. F\! would be convex over all q_{1}.

By the convexity of \! F with respect to \! q_{1} \!, there would be a unique q_{1} that globally maximizes F. The first derivative of F\! with respect to q_{1}\! should be 0 for that particular \! q_{1}. In other words:

\displaystyle \frac{k}{r^{2}}\, (q - 2\, q_{1}) = 0<em>.</em>

2\, q_{1} = q.

q_{1} = q / 2.

In other words, the force between the two point charges would be maximized when the charge is evenly split:

\begin{aligned} \frac{q}{q_{1}} &= \frac{q}{q / 2} = 2\end{aligned}.

3 0
3 years ago
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