Answer:
The magnitude of the magnetic torque on the coil is 1.98 A.m²
Explanation:
Magnitude of magnetic torque in a flat circular coil is given as;
τ = NIASinθ
where;
N is the number of turns of the coil
I is the current in the coil
A is the area of the coil
θ is the angle of inclination of the coil and magnetic field
Given'
Number of turns, N = 200
Current, I = 7.0 A
Angle of inclination, θ = 30°
Diameter, d = 6 cm = 0.06 m
A = πd²/4 = π(0.06)²/4 = 0.002828 m²
τ = NIASinθ
τ = 200 x 7 x 0.002828 x Sin30
τ = 1.98 A.m²
Therefore, the magnitude of the magnetic torque on the coil is 1.98 A.m²
Answer: You didn't put what a, b, c, or d is so how is anyone supposed to know what the answer is?
The magnitude of the magnetic field inside the solenoid is 3.4×10^(-4) T.
To find the answer, we need to know about the magnetic field inside the solenoid.
<h3>What's the expression of magnetic field inside a solenoid?</h3>
- Mathematically, the expression of magnetic field inside the solenoid= μ₀×n×I
- n = no. of turns per unit length and I = current through the solenoid
<h3>What's is the magnetic field inside the solenoid here?</h3>
- Here, n = 290/32cm or 290/0.32 = 906
I= 0.3 A
- So, Magnetic field= 4π×10^(-7)×906×0.3 = 3.4×10^(-4) T.
Thus, we can conclude that the magnitude of the magnetic field inside the solenoid is 3.4×10^(-4) T.
Learn more about the magnetic field inside the solenoid here:
brainly.com/question/22814970
#SPJ4
1. Velocity
2. Time
3. Idk
4. Idk
5. D. I think it may be A. but I think D.
