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disa [49]
3 years ago
14

A 600-turn solenoid, 25 cm long, has a diameter of 2.5 cm. A 14-turn coil is wound tightly around the center of the solenoid. If

the current in the solenoid increases uniformly from 0 to 5.0 A in 0.60 s, what will be the induced emf in the short coil during this time
Physics
1 answer:
Delvig [45]3 years ago
8 0

Answer:

The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

Explanation:

The magnetic field at the center of the solenoid is given by;

B = μ(N/L)I

Where;

μ is permeability of free space

N is the number of turn

L is the length of the solenoid

I is the current in the solenoid

The rate of change of the field is given by;

\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s

The induced emf in the shorter coil is calculated as;

E = NA\frac{\delta B}{\delta t}

where;

N is the number of turns in the shorter coil

A is the area of the shorter coil

Area of the shorter coil = πr²

The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m

Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²

E = NA\frac{\delta B}{\delta t}

E = 14 x 0.000491 x 0.02514

E = 1.728 x 10⁻⁴ V

Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

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Two balls, each with a mass of 0.5 kg, collide on a pool table. Is the law of conservation of momentum satisfied in the collisio
Dmitriy789 [7]

During the collision between two balls on the pool table there is no external force along the line of collision between them

Since there is no external force on it so here we can say

F = 0 = \frac{\Delta P}{\Delta t}

here we have

\Delta P = 0

so we can say

P_i = P_f

since there is no external force so we can say during the collision the momentum of two balls will remain conserved

7 0
3 years ago
Read 2 more answers
A rectangular copper strip 1.5cm wide and 0.10cn thick carries a current of 5.0A. Find the Hall voltage for a 1.2T magnetic fiel
Scrat [10]

Answer:

4.4345× 10^-7V

Explanation:

The computation of the half voltage for a 1.2T magnetic field applied is shown below

The volume of one mole of copper is

v = m ÷p

= 63.5 ÷ 8.92

= 7.12cm

Now the density of free electrons in copper is

n = Na ÷ V

= 6.02 × 10^23 ÷ 7.12

= 8.456× 10^28/m^3

Now the half voltage is

= IB ÷ nqt

= (5 × 1.20) ÷ (8.456× 10^28 × 1.6 × 10^-19 × 0.1× 10^-2)

= 4.4345× 10^-7V

7 0
2 years ago
If a 6V battery is connected to a light bulb whose resistance is 55,000Ω How much current will flow in the circuit?
notka56 [123]

Answer:

Current, I = 0.000109 Amps

Explanation:

Given the following data;

Voltage = 6V

Resistance = 55,000 Ohms

To find the current flowing through the circuit;

Ohm's law states that at constant temperature, the current flowing in an electrical circuit is directly proportional to the voltage applied across the two points and inversely proportional to the resistance in the electrical circuit.

Mathematically, Ohm's law is given by the formula;

V = IR

Where;

V represents voltage measured in voltage.

I represents current measured in amperes.

R represents resistance measured in ohms.

Making current the subject of formula, we have;

I = \frac {V}{R}

Substituting into the formula, we have;

I = \frac {6}{55000}

Current, I = 0.000109 Amps

5 0
3 years ago
ball of mass 0.4 kg is attached to the end of a light stringand whirled in a vertical circle of radius R = 2.9 m abouta fixed po
Levart [38]

Answer:

6.046N

Explanation:

The net force exerted on the mass is the sum of tension force and the external force of gravity.

F_n_e_t=F_g+F_t

F_t is the tension force.F_g=9.8N/kg is the force of gravity.

F_n_e_t=ma_c=mv^2/r\\

where r is the rope's radius from the fixed point.

From the net force equation above:

F_t=F_n_e_t-F_g\\=mv^2/r-mg\\=0.4\times(8.5^2/2.9)-0.4\times9.8\\=6.046N

Hence the tension force is 6.046N

8 0
3 years ago
How will the motion of the arrow change after it leaves the bow?
Pavel [41]

The string moves to the right, as it restores its original position with the median plane of the bow. As a result, the string "pulls" on the arrow with a force F2. 2. The tip of the arrow T moves slightly to the left.

pls thank me and brainliest me

4 0
3 years ago
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