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Correct question:
A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the center of the solenoid is 2.8 x 10⁻² T, what is the number of turns per meter for this solenoid?
Answer:
the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.
Explanation:
Given;
length of solenoid, L= 0.35 m
diameter of the solenoid, d = 0.04 m
current through the solenoid, I = 5.0 A
magnetic field in the center of the solenoid, 2.8 x 10⁻² T
The number of turns per meter for the solenoid is calculated as follows;

Therefore, the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.
Answer:
x ’= 368.61 m, y ’= 258.11 m
Explanation:
To solve this problem we must find the projections of the point on the new vectors of the rotated system θ = 35º
x’= R cos 35
y’= R sin 35
The modulus vector can be found using the Pythagorean theorem
R² = x² + y²
R = 450 m
we calculate
x ’= 450 cos 35
x ’= 368.61 m
y ’= 450 sin 35
y ’= 258.11 m