You have a 2m long wire which you will make into a thin coil with N loops to generate a magnetic field of 3mT when the current i
1 answer:
Answer:
<em>radius of the loop = 7.9 mm</em>
<em>number of turns N ≅ 399 turns</em>
Explanation:
length of wire L= 2 m
field strength B = 3 mT = 0.003 T
current I = 12 A
recall that field strength B = μnI
where n is the turn per unit length
vacuum permeability μ = = 1.256 x 10^-6 T-m/A
imputing values, we have
0.003 = 1.256 x 10^−6 x n x 12
0.003 = 1.507 x 10^-5 x n
n = 199.07 turns per unit length
for a length of 2 m,
number of loop N = 2 x 199.07 = 398.14 ≅ <em>399 turns</em>
since there are approximately 399 turns formed by the 2 m length of wire, it means that each loop is formed by 2/399 = 0.005 m of the wire.
this length is also equal to the circumference of each loop
the circumference of each loop =
0.005 = 2 x 3.142 x r
r = 0.005/6.284 = = 0.0079 m =<em> 7.9 mm</em>
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Answer:
The magnitude of electric force is
Explanation:
Coulomb's Law:
The force of attraction or repletion is
- directly proportional to the products of charges i.e
- inversely proportional to the square of distance i.e
[ k is proportional constant=9×10⁹N m²/C²]
There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C
Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C
and F₂ force be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C
Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.
If we draw a line from q₁ to Q .
The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.
Let hypotenuse = r
Therefore,
we know,
Total force
[ r=5]
N
The magnitude of electric force is
F = Magnetic Force
B = Magnetic Field
V = Velocity
*The vectors from the photo you get doing the left-hand rule.
The magnetic force is always perpendicular to the magnetic field.
And as told in the statement, the electron is moving perpendicular to a magnetic field, that is, the Velocity forms an 90 degree angle / Right angle with the magnetic field.
The formula to find the Magnetic Force is:
Where "q" is the Charge and the sin theta is the angle formed by the Velocity and Magnetic Field, in this case it's 90°. Sin 90° = 1.
Newton (N) = C x m/s x T = (C x m x T)/s
B = Magnetic Field
V = Velocity
*The vectors from the photo you get doing the left-hand rule.
The magnetic force is always perpendicular to the magnetic field.
And as told in the statement, the electron is moving perpendicular to a magnetic field, that is, the Velocity forms an 90 degree angle / Right angle with the magnetic field.
The formula to find the Magnetic Force is:
Where "q" is the Charge and the sin theta is the angle formed by the Velocity and Magnetic Field, in this case it's 90°. Sin 90° = 1.
Newton (N) = C x m/s x T = (C x m x T)/s