Answer:
Explanation:
a ) No of turns per metre
n = 450 / .35
= 1285.71
Magnetic field inside the solenoid
B = μ₀ n I
Where I is current
B = 4π x 10⁻⁷ x 1285.71 x 1.75
= 28.26 x 10⁻⁴ T
This is the uniform magnetic field inside the solenoid.
b )
Magnetic field around a very long wire at a distance d is given by the expression
B = ( μ₀ /4π ) X 2I / d
= 10⁻⁷ x 2 x ( 1.75 / .01 )
= .35 x 10⁻⁴ T
In the second case magnetic field is much less. It is due to the fact that in the solenoid magnetic field gets multiplied due to increase in the number of turns. In straight coil this does not happen .
His average speed is (35m/45s) = 7/9 meters per second.
His average velocity is (30m W + 5m E) / (45s) = 25 m/s West .
Answer:
An electromagnet is made by forming a coil around a soft iron bar (known here as the metal) such as a nail or screw and connect with an insulated copper wire (known here as the electric current conductor) the ends of the wound copper is then connected separately to the positive and negative terminals of a battery (known here as the source of electric current)
The north seeking needle of the magnetic compass will move away when brought close to the north pole of the formed electromagnet which can then be labelled N
The magnetic compass needle will be attracted to the south pole of the electromagnet which can then be labelled S
Explanation:
An electromagnet is an electric powered magnet that is formed (temporarily) by the perpendicular movement of electric current with respect to a metal core
The magnitude and the poles of an electromagnet can be changed by changing the magnitude and the direction of flow of the electric current respectively.
Usually, in culturing of the bacteria we have a slant and then portion f it is transferred to the agar plate. The growth characteristics are more useful in the agar plates because it is where we really do the observation because bacteria in slants are still to be transferred in the agar plates.
Answer:
8N and 32N
Explanation:
Given that a light board, 10 m long, is supported by two sawhorses, one at one edge of the board and a second at the midpoint. A small 40-N weight is placed between the two sawhorses, 3.0 m from the edge and 2.0 m from the center.
To calculate the forces that are exerted by the sawhorses on the board, we must consider the equilibrium of forces acting on the board.
Let the two upward forces produce by the saw horses be P1 and P2
Assuming that the weight is negligible
Sum of the upward forces = sum of the downward forces.
P1 + P2 = 40 ....... (1)
Also, the sum of the clockwise moment = sum of the anticlockwise moments.
Let's assume that the board is uniform. The weight will act at the centre.
Taking moment at the centre:
P1 × 5 + 40 × 2 = 0
P1 = 40 / 5
P1 = 8N
Substitute P1 into equation 1
8 + P2 = 40
P2 = 40 - 8
P2 = 32N