Question

DQuestion 21 1 pts 53.13 2 Figure F3 Current-carrying square coil in magnetic field A square coil has N-40 turns of copper wire, each of side a-5.0 cm. The current in the coil is 1-4.0 A. The coil is placed in a uniform magnetic field B-16 T as shown in Figure F3. The plane of the coil makes an angle of 53.13 with the magnetic field. The coil is free to rotate about the y axis. Calculate the area of the coil [in m2]. 0.1 DQuestion 22 1 pts Calculate u, the magnetic moment [in A-m2] of the coil. DQuestion 23 1 pts Calculate ф the angle [in degrees] between the magnetic moment and the magnetic field B.

Answer 1

"Given that equation is rightdoublearrow The given equation is the magnetic moment to calculate rightdoublearrow The given magnet moment (u) = IA Here I = 40 A a = 50 cm: = 5 times 10^-2 m u = IA = 4.0 times 25 times 10^-4 m^2 A = 100 times 10^-4 m^2 = 10^-2 Am^2 therefore u = 0.1 Am^2 rightdoublearrow is the direction to the perpendicular to the plane to the loop equation The given equation is the angle b/w magnetic moment ""mu"" and the magnet field B emptyset = 90 - 53.13 degree = 26.87 degree \r\n The given equation is T Torque the [N-m] explained the coil by the magnetic tau rightarrow = mu rightarrow times Bra = Vertical-bar u rightarrow Vertical-bar Vertical-bar B rightarrow Vertical-bar sin emptyset tau = 01 times 1.6 times sin (36.87 degree)Nm tau = 0.0096Nm therefore Hence the equation's are proved."