One man pushes on the front of a cart while

A coil of 40 turns is wrapped around a long solenoid of cross-sectional area 7.5×10−3m2. The solenoid is 0.50 m long and has 500

defon

To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.

This definition is described in the following equation as,

$M = \frac{\mu_0 N_1 N_2A_1}{l_1}$

Where,

$\mu =$ permeability of free space

$N_1 =$ Number of turns in solenoid 1

$N_2 =$ Number of turns in solenoid 2

$A_1=$ Cross sectional area of solenoid

l = Length of the solenoid

Part A )

Our values are given as,

$\mu_0 = 4\pi *10^{-7}H/m$

$N_1 = 500$

$N_2 = 40$

$A = 7.5*10^{-4}m^2$

$l = 0.5m$

Substituting,

$M = \frac{\mu_0 N_1 N_2A_2}{l_1}$

$M = \frac{(4\pi *10^{-7})(500)(40)(7.5*10^{-4})}{0.5}$

$M = 3.77*10^{-4}H$

PART B) Considering that many of the variables remain unchanged in the second solenoid, such as the increase in the radius or magnetic field, we can conclude that mutual inducantia will appear the same.

8 0

4 years ago

What are the advantages and disadvantages of renewable and non-renewable energy?

bogdanovich [222]

Renewable energy

Advantages :-

1. Easily regenerate

2. Boost economic growth

3. Easily available

4. Support environment

5. Low maintenance cost

Disadvantages :-

1. Weather dependency

2. High installation cost

3. Noise caused by wind energy

4. Fluctuation problem (solar)

5. Intermittency issue (wind)

Non-renewable energy

Advantages :-

1. Concentrated energy source

2. Reliable energy source

3. Can be built anywhere

4. No radioactive waste

Disadvantages :-

1. Produces greenhouse gases

2. Contributes to global warming

3. Produces acid rain

4. Harmful to environment when they are burnt

I hope this helps.....

5 0

3 years ago

Two long, parallel wires carry currents of different magnitudes. If the amount of current in one of the wires is doubled, what h

Ulleksa [173]

Answer:

The magnitude of the force that each wire exerts on the other will increase by a factor of two.

Explanation:

force on parallel current carrying wire, F = BILsinθ

where;

B is the strength of the magnetic field

L is the length of the wire

I is the magnitude of current on the wire

θ is the angle of inclination of the wire

Assuming B, L and θ is constant, then F ∝ I

F = kI

$\frac{F_1}{I_1} = \frac{F_2}{I_2}$

When the amount of current is doubled in one of the wires, lets say the second wire;

$\frac{F_1}{I_1} = \frac{F_2}{2I_1} \\\\F_2 = \frac{2F_1I_1}{I_1} \\\\F_2 =2F_1$

Also, if will double the amount of current on the first wire, then

F₁ = 2F₂

Therefore, the magnitude of the force that each wire exerts on the other will increase by a factor of two.

3 0

4 years ago

A particle with a charge of 5 × 10–6 C and a mass of 20 g moves uniformly with a speed of 7 m/s in a circular orbit around a sta

creativ13 [48]

Answer:

r = 0.22m

Explanation:

To find the radius of the circular trajectory, you first take into account that the centripetal force of the charged particle, is equal to the electric force between the particle that is moving and the particle at the center of the orbit.

Then, you have:

$F_c=F_e=ma_c$ (1)

m: mass of the particle = 20g = 20*10-3 kg

ac: centripetal acceleration = ?

q: charge of the particle = 5*10^-6C

Fe: electric force between the charges

The electric force is given by:

$F_e=k\frac{qq'}{r^2}$ (2)

r: radius of the orbit

q': charge of the particle at the center of the orbit = -5*10^-6C

Furthermore, the centripetal acceleration is:

$a_c=\frac{v^2}{r}$ (3)

v: speed of the particle = 7m/s

You replace the expressions (2) and (3) in the equation (1) and solve for r:

$k\frac{qq'}{r^2}=m\frac{v^2}{r}\\\\r=\frac{kqq'}{mv^2}$

Finally, you replace the values of all parameters in the previous expression:

$r=\frac{(8.98*10^9Nm^2/C^2)(5*10^{-6}C)(5*10^{-6}C)}{(20*10^{-3}kg)(7m/s)^2}\\\\r=0.22m$

The radius of the circular trajectory is 0.22m

5 0

3 years ago

A beam of light converging to the point of 10 cm is incident on the lens. find the position of the point image if the lens has a

Verizon [17]

Answer:

beam of light converges to a point A. A lens is placed in the path of the convergent beam 12 cm from P.

To find the point at which the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm

Solution:

As per the given criteria,

the the object is virtual and the image is real (as the lens is placed in the path of the convergent beam)

(a) lens is a convex lens with

focal length, f=20cm

object distance, u=12cm

applying the lens formula, we get

f

1

=

v

1

−

u

1

⟹

v

1

=

f

1

+

u

1

⟹

v

1

=

20

1

+

12

1

⟹

v

1

=

60

3+5

⟹v=7.5cm

Hence the image formed is real, at 7.5cm from the lens on its right side.

(b) lens is a concave lens with

focal length, f=−16cm

object distance, 12cm

applying the lens formula, we get

f

1

=

v

1

−

u

1

⟹

v

1

=

f

1

+

u

1

⟹

v

1

=

−16

1

+

12

1

⟹

v

1

=

48

−3+4

⟹v=48m

Hence the image formed is real, at 48 cm from the lens on the right side.

6 0

3 years ago