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faltersainse [42]

3 years ago

5

5)

1 answer:

miskamm [114]3 years ago

4 0

B . I hope this is right

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What is the magnitude of the magnetic field at a point midway between them if the top one carries a current of 19.5 A and the bo

Phantasy [73]

Answer:

The magnetic field will be $\large{\dfrac{1.4 \times 10^{-4}}{d}} T$ , '2d' being the distance the wires.

Explanation:

From Biot-Savart's law, the magnetic field ( $\large{\overrightarrow{B}}$ ) at a distance ' $r$ ' due to a current carrying conductor carrying current ' $I$ ' is given by

$\large{\overrightarrow{B} = \dfrac{\mu_{0}I}{4 \pi}} \int \dfrac{\overrightarrow{dl} \times \hat{r}}{r^{2}}}$

where ' $\overrightarrow{dl}$ ' is an elemental length along the direction of the current flow through the conductor.

Using this law, the magnetic field due to straight current carrying conductor having current ' $I$ ', at a distance ' $d$ ' is given by

$\large{\overrightarrow{B}} = \dfrac{\mu_{0}I}{2 \pi d}$

According to the figure if ' $I_{t}$ ' be the current carried by the top wire, ' $I_{b}$ ' be the current carried by the bottom wire and ' $2d$ ' be the distance between them, then the direction of the magnetic field at 'P', which is midway between them, will be perpendicular towards the plane of the screen, shown by the $\bigotimes$ symbol and that due to the bottom wire at 'P' will be perpendicular away from the plane of the screen, shown by $\bigodot$ symbol.

Given $\large{I_{t} = 19.5 A}$ and $\large{I_{B} = 12.5 A}$

Therefore, the magnetic field ( $\large{B_{t}}$ ) at 'P' due to the top wire

$B_{t} = \dfrac{\mu_{0}I_{t}}{2 \pi d}$

and the magnetic field ( $\large{B_{b}}$ ) at 'P' due to the bottom wire

$B_{b} = \dfrac{\mu_{0}I_{b}}{2 \pi d}$

Therefore taking the value of $\mu_{0} = 4\pi \times 10^{-7}$ the net magnetic field ( $\large{B_{M}}$ ) at the midway between the wires will be

$\large{B_{M} = \dfrac{4 \pi \times 10^{-7}}{2 \pi d} (I_{t} - I_{b}) = \dfrac{2 \times 10^{-7}}{d} = \dfrac{41.4 \times 10 ^{-4}}{d}} T$

5 0

3 years ago

What is the relationship between elevation and energy (think about the same can dropped from different heights)?

MA_775_DIABLO [31]

Elevation would be showing you what height you are at, energy would be like what force your putting into the object.

8 0

3 years ago

Read 2 more answers

1.A river flowing steadily at a rate of 240 m3/s is considered for hydroelectric power generation. It is determined that a dam c

SIZIF [17.4K]

Answer:

the power that can be generated by the river is 117.6 MW

Explanation:

Given that;

Volume flow rate of river v = 240 m³/s

Height above the lake surface a h = 50 m

Amount of power can be generated from this river water after the dam is filled = ?

Now the collected water in the dam contains potential energy which is used for the power generation,

hence, total mechanical energy is due to potential energy alone.

$E_{mech}$ = m(gh)

first we determine the mass flow rate of the fluid m

m = p×v

where p is density ( 1000 kg/m³

so we substitute

m = 1000kg/m³ × 240 m³/s

m = 240000 kg/s

so we plug in our values into ( $E_{mech}$ = m(gh) kJ/kg )

$E_{mech}$ = 240000 × 9.8 × 50

$E_{mech}$ = 117600000 W

$E_{mech}$ = 117.6 MW

Therefore, the power that can be generated by the river is 117.6 MW

4 0

3 years ago

If a 5-L balloon at 25 degrees celsius were gently heated to 30 degrees celsius, what new volume would the balloon have? Show al

WITCHER [35]

Answer: 5.08 L.

Explation down below

4 0

2 years ago

A 800N student does a handstand with both hands at an angle of 15 degrees from the vertical what is the force on each hand

jeyben [28]

here we can say that net force on the student vertically upwards will be counter balance by his weight downwards

Let net force F is exerted by each hand

so here we will have

$2Fcos15 = W$

$2F cos15 = 800$

$F = \frac{800}{2 cos15}$

$F = 414.1 N$

so the force exerted by each hand will be 414.1 N

5 0

3 years ago