Answer:
the length of the sides s is [tex]s = 1.998 \ cm[/tex]
Explanation:
From the question we are told that
The number of turns is [tex]N = 60 \ turn[/tex]
The magnetic field is [tex]B = 1.20 \ T[/tex]
The angle the loop makes with the x-axis [tex]\theta = 15 ^o[/tex]
The current flowing through the loop is [tex]I = 2.50 A[/tex]
The magnitude of the torque is [tex]\tau = 0.0186 \ N[/tex]
the length of the sides of the square is [tex]s[/tex]
Generally, we can represent the torque magnitude as
[tex]\tau = N I A B sin \theta[/tex]
Where A is the area of the square which is mathematically represented as
[tex]A = s^2[/tex]
Substituting this into the formula for torque
[tex]\tau = N I s^2 B sin \theta[/tex]
making s the subject
[tex]s = \sqrt{\frac{\tau }{NIB sin \theta } }[/tex]
Substituting values
[tex]s = \sqrt{\frac{0.0186}{(60) * (2.50) * (1.20) * (sin (15))} }[/tex]
[tex]s = 0.01998 m[/tex]
Converting to centimeters
[tex]s = 0.01998 * 100[/tex]
[tex]s = 1.998 \ cm[/tex]
Answer:
2 cm
Explanation:
To fins the lengths of the sides of the loop you use the following formula for the calculation of the torque experienced by the loop in a magnetic field:
[tex]\tau=NiABsin\theta[/tex]
N: turns of the loop = 60
i: current in the loop = 2.50A
A: area of the loop = s*s
B: magnitude of the magnetic field = 1.20T
θ: angle between the plane of the lop and the direction of B = 15°
BY replacing the values of the torque and the other parameters in (1) you can obtain the area of the loop:
[tex]A=\frac{\tau}{NiBsin\tetha}=\frac{0.0186Nm}{(60)(2.5A)(1.2T)sin15\°}=3.99*10^{-4}m^2[/tex]
but the area is s*s:
[tex]A=\sqrt{s}=\sqrt{3.99*10^{-4}m^2}=0.019m\approx2cm[/tex]
hence, the sides of the square loop have a length of 2cm
