Answer:
The average magnitude of magnetic field B= 0.0433/ d Tesla
(You have not provided length of side of loop, so if you divide this value by length you will get value of magnetic field.)
Explanation:
Induced emf
where B= magnetic field
d= breadth of rectangular piece
V= velocity with which the rectangular piece = o.o6m/s
n= no of turns = 10
EMF = 26mV
since d (breadth of the frame) is not given, I will use it as a variable
EMF= n×B×d×V ------------------(1) (EMF induced due to multiple turns)
From eq 1, we get
B= (EMF)/(n d V)
B= (26 X 0.001) / (10 d 0.06)
B= 0.0433/ d Tesla
There is a ner force of 15 N allowing Levi and Clara to mobe the box.
<h3>Question:</h3>
A 2.0-cm length of wire centered on the origin carries a 20-A current directed in the positive y direction. Determine the magnetic field at the point x = 5.0m on the x-axis.
<h3>
Answer:</h3>
1.6nT [in the negative z direction]
<h2>
Explanation:</h2>
The magnetic field, B, due to a distance of finite value b, is given by;
B = (μ₀IL) / (4πb
) -----------(i)
Where;
I = current on the wire
L = length of the wire
μ₀ = magnetic constant = 4π × 10⁻⁷ H/m
From the question,
I = 20A
L = 2.0cm = 0.02m
b = 5.0m
Substitute the necessary values into equation (i)
B = (4π × 10⁻⁷ x 20 x 0.02) / (4π x 5.0
)
B = (10⁻⁷ x 20 x 0.02) / (5.0
)
B = (10⁻⁷ x 20 x 0.02) / (5.0
)
B = (10⁻⁷ x 20 x 0.02) / (25.0)
B = 1.6 x 10⁻⁹T
B = 1.6nT
Therefore, the magnetic field at the point x = 5.0m on the x-axis is 1.6nT.
PS: Since the current is directed in the positive y direction, from the right hand rule, the magnetic field is directed in the negative z-direction.