Ask question

Ask question

All categories

3 years ago

15

A wire with a current of 3.40 A is to be formed into a circular loop of one turn. If the required value of the magnetic field at

the center of the loop is 20 µT, what is the required radius?

1 answer:

Sloan [31]3 years ago

7 0

Answer:

0.107 m

Explanation:

The magnetic field at the center of a current-carrying loop is given by

$B=\frac{\mu_0 I}{2r}$

where

$\mu_0$ is the vacuum permeability

I is the current

r is the radius of the loop

In this problem we have

I = 3.40 A is the current in the loop

$B=20 \mu T=20\cdot 10^{-6}T$ is the magnetic field at the centre of the loop

So, solving the formula for r we find

$r=\frac{\mu_0 I}{2B}=\frac{(12.56\cdot 10^{-7} H/m)(3.40 A)}{2(20\cdot 10^{-6} T)}=0.107 m$

You might be interested in

cylindrical container is to be constructed to be open at the top with a volume of 27π cubic meters using the least amount of mat

Llana [10]

Answer:

radius comes out to be 3 m

height of the cylinder comes out to be 3m

Explanation:

given

volume of cylinder = 27π m³

π r² h = 27π

r² h = 27.............(1)

surface area of cylinder open at the top

S = 2πrh + π r²

$S = 2\pi \dfrac{27}{r} + \pi r^2$

$\frac{\mathrm{d} s}{\mathrm{d} r}=\frac{\mathrm{d}}{\mathrm{d} r} (2\pi \dfrac{27}{r} + \pi r^2)$

$\frac{\mathrm{d} s}{\mathrm{d} r}=54\pi \dfrac{-1}{r^2}+2\pi r$

$\frac{\mathrm{d} s}{\mathrm{d} r}=0$

for least amount of material requirement.

$\dfrac{54\pi }{r^2} = 2\pi r\\r=3m$

hence radius comes out to be 3 m

for height put the value in the equation 1

so, height of the cylinder comes out to be 3m

3 0

3 years ago

Read 2 more answers

On the planet Xenophous a 1.00 m long pendulum on a clock has a period of 1.32 s. What is the free fall acceleration on Xenophou

Vilka [71]

The period of the pendulum is given by the following equation

T = 2π * sqrt (L/g)

Where g is the gravity (free fall acceleration)

L is the longitude of the pendulum

T is the period.

We find g.............> (T /2π)^2 = L/g

g = L/(T /2π)^2...........> g = 22.657 m/s^2

3 0

3 years ago

My Notes (a) A 41 Ω resistor is connected in series with a 6 µF capacitor and a battery. What is the maximum charge to which thi

Lilit [14]

Explanation:

The give data is as follows.

C = 6 $\mu F$ = $6 \times 10^{-6} F$

V = 6 V

Now, we know that the relation between charge, voltage and capacitor for series combination is as follows.

Q = CV

= $6 \times 10^{-6} F \times 6 V$

= $36 \times 10^{-6} C$

or, = 36 $\mu C$

Thus, we can conclude that maximum charge of the given capacitor is 36 $\mu C$ .

8 0

3 years ago

Read 2 more answers

A 2.6 mm -diameter sphere is charged to -4.5 nC . An electron fired directly at the sphere from far away comes to within 0.37 mm

motikmotik

Answer: $1.96\times 10^{8}\ m/s$

Explanation:

Given

Diameter of sphere is $d=2.6\ mm\quad \quad[r=1.3 mm]$

Charge on the sphere is $Q=-4.5\ nC$

Nearest distance electron can reach to sphere is $d=0.37\ mm$

Here, kinetic energy of electron is converted into electrostatic energy between the two i.e.

$\Rightarrow \dfrac{1}{2}mv^2=\dfrac{kQq}{d}\\\\\Rightarrow \dfrac{1}{2}\times 9.1\times 10^{-31}\times v^2=\dfrac{9\times 10^9\times 4.5\times 10^{-9}\times 1.6\times 10^{-19}}{(3.7\times 10^{-4})}\\\\\Rightarrow v^2=3.8491\times 10^{16}\\\Rightarrow v=1.96\times 10^{8}\ m/s$

Thus, the initial speed of electron is $1.96\times 10^{8}\ m/s$ .

7 0

3 years ago

A horizontal force of 40N is needed to pull a 60kg box across the horizontal floor at which coefficient of friction between floo

Mazyrski [523]

Answer:

Coefficient of friction is $0.068$ .

Work done is $320~J$ .

Explanation:

Given:

Mass of the box ( $m$ ): $60$ kg

Force needed ( $F$ ): $40$ N

The formula to calculate the coefficient of friction between the floor and the box is given by

$F=\mu mg...................(1)$

Here, $\mu$ is the coefficient of friction and $g$ is the acceleration due to gravity.

Substitute $40$ N for $F$ , $60$ kg for $m$ and $9.80$ m/s² for $g$ into equation (1) and solve to calculate the value of the coefficient of friction.

$40 N=\mu\times60 kg\times9.80 m/s^{2} \\~~~~~\mu=\frac{40 N}{60 kg\times9.80 m/s^{2}}\\~~~~~~~=0.068$

The formula to calculate the work done in overcoming the friction is given by

$W=Fd..........................(2)$

Here, $W$ is the work done and $d$ is the distance travelled.

Substitute $40$ N for $F$ and $8 m$ for $d$ into equation (2) to calculate the work done.

$W=40~N\times8~m\\~~~~= 320~J$

8 0

2 years ago