A long solenoid that has 1000 turns uniformly distributed over a length of 0.400 m produces a magnetic field of magnitude 1.00 ×
1 answer:
0.03185A current is required in the windings for that to occur.
- Solenoids are essentially coils of cord. These generate a magnetic subject which strives a pressure over a steel element.A solenoid is a fundamental time period for a coil of cord that we use as an electromagnet.
- We additionally consult with the tool which could convert electric power into mechanical power as a solenoid.
- Actually it generates a magnetic subject for developing linear movement from the electrical cutting-edge. With using a magnetic subject.
- Magnetic field at the centre of a solenoid of length 'L' having 'N' number of turns with a current 'I' is given by
...(1)
It is given that 1000 turns uniformly distributed over a length of 0.400 m produces a magnetic field of magnitude 1.00 ×10⁻⁴T .
Putting above values in equation (1) , we get
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<h2>Answer:</h2>
(a) standard deviation = σ = 4.9996
(b) variance = σ² = 24.996
<h2>Explanation:</h2><h2 /><em>Given frequency table (find attached as Table 1);</em>
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(a) To find the sample standard deviation and sample variance, follow these steps;
<em>i. Calculate the mid-point c for each group by using the mid-point formula;</em>
c = (lower bound + upper bound) / 2
=> c = (6.51 + 8.50) / 2 = 7.505
=> c = (8.51 + 10.50) / 2 = 9.505
=> c = (10.51 + 12.50) / 2 = 11.505
=> c = (12.51 + 14.50) / 2 = 13.505
=> c = (14.51 + 16.50) / 2 = 15.505
<em>So the new table becomes (find attached as Table 2);</em>
<em>ii. Calculate the total number of samples (n) which is the sum of all the frequencies.</em>
n = 50+18+42+20+46
n = 176
<em>iii. Calculate the mean (M)</em>
This is done by first multiplying the midpoints by the corresponding frequencies and then dividing the result by the total number of samples (n).
M = [(7.505 x 50) + (9.505 x 18) + (11.505 x 42) + (13.505 x 20) + (15.505 x 46)] / 176
M = [375.25 + 171.09 + 483.21 + 270.1 + 713.23] / 176
M = [2012.88] / 176
M = 11.44
<em>iv. Find the variance (σ²);</em>
The variance is calculated using the following formula
σ² = [Σ(f x c²) - (n x M²)] / (n - 1) ------------(i)
Where;
f = frequency of each boundary data point
<em>=> Let's first calculate </em>Σ(f x c²).
This is done by finding the sum of the product of the frequency (f) of each boundary point and the square of their corresponding mid-points(c)
Σ(f x c²) = [(50 x 7.505²) + (18 x 9.505²) + (42 x 11.505²) + (20 x 13.505²) + (46 x 15.505²)]
Σ(f x c²) = [(2816.25125) + (1626.21045) + (5559.33105) + (3647.7005) + (11058.63115)]
Σ(f x c²) = 24708.1244
<em>=> Now calculate (n x M²)</em>
n x M² = 176 x 11.44²
n x M² = 23033.7536
<em>=> Now substitute these values into equation (i) to calculate the variance</em>
σ² = [Σ(f x c²) - (n x M²)] / (n - 1)
σ² = [24708.1244 - 23033.7536] / (176 - 1)
σ² = [4374.3708] / (175)
σ² = 24.996
Therefore, the variance is 24.996
<em>v. Find the standard deviation (σ)</em>
The standard deviation is the square root of the variance. i.e
σ = √σ²
σ = √24.996
σ = 4.9996
Therefore, the standard deviation is 4.9996