These questions refer to the expression for the magnetic field at the center of

Question

These questions refer to the expression for the magnetic field at the center of the square solenoid. Assume: for all coils, each side of the square a = 2.27 cm and the length of the solenoid L = 58.3 cm.
NOTE: Everything must be in MKS units!
NOTE: o = 1.26x10-6 in MKS units (determined in question 1).

a) In procedure 1, you will plot Bmax vs. N, the number of turns of the coil. Suppose you set the current of I = 1.55 A, and vary the number of turns of the coil (with a and L the same for each coil). Find the slope of this straight line.
slope = ________T/turn

b) In procedure 2, you will plot Bmax vs. I. Suppose you use a coil with 214 turns, and vary the current I. Find:
- n, the number of turns per length: n = ______ turns/m
- the slope of this straight line: slope = _________T/A

Explanation / Answer

Given,

each side of the square a = 2.27 cm

the length of the solenoid L = 58.3 cm.

the current of I = 1.55 A

Maximum magnetic field inside a solenoid is

Bmax = o * N * I / L

=1.26 * 10-6 * N * I /0.583

= 2.161 * 10-6 * N * I

a) I=1.55 A

Bmax =2.161* 10-6 * N * 1.55

=3.349 * 10-6 T

y axis-1 Unit=10-3 T

Slope of the graphm=(3.349 * 10-3 - 0) / (1000 - 0)

=3.349 * 10-6 T/turn

b) Number of turns per unit lengthn=N/L

=214 / 0.583

=367.06 /m

Bmax = 1.26 * 10-6 * 367.06* I

=0.000462 * IT

n = 214/0.583

=367.06 turns/s

the number of turns per length n=367.06 turns/s

slope = 1.26*10^-6*367.06

= 0.000462495 T/A

the slope of this straight line is 0.000462495 T/A

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