The magnitude of the magnetic field at a perpendicular distance r from a long st
ID: 1607865 • Letter: T
Question
The magnitude of the magnetic field at a perpendicular distance r from a long straight wire carrying a current is given by: B = mu_o i/2 pi r where mu_0 = 4 pi times 10^-7 Tm/A is the permeability of free space The direction of this field can be determined by the right-hand rule. A long wire carries a current i_1 and a rectangular loop carries current i_2 are placed in the same plane as shown below. Assume the following values: i_1 = 12.0 A, i_2 = 33.0 A, a = 1.25 cm, b = 7.50 cm, and L = 15.0 cm. (A) Find the direction and the magnitude of the net magnetic force on the loop due to the current i_1. (B) Find the net torque about the axis cc' as shown in the figure. (Axis cc' lies in the same plane of the loop, and is parallel to the current i_1. It lies midway between loop sides AB and CD as shown below)Explanation / Answer
(A) Field at the location of AB part,
B = u0 i1 / 2 pi a
and F = IL X B
as IL and B are perpendicular, F = I L B
F1 = u0 i1 i2 L / (2 pi a)
F1 = (4pi x 10^-7 x 12 x 33) / (2 x pi x 0.0125) towrads the wire
F1 = 6.34 x 10^-3 N to the left
similarly force on CD,
F2 = u0 i1 i2 L / (2 pi (a + b))
F2 = (4pi x 10^-7 x 12 x 33) / (2 x pi x (0.0125 + 0.0750))
F2 = 0.905 x 10^-3 N to the right
forces on AD and BC due to wire will be same in magnitude but opposite on wire.
F_AD + F_BC = 0
Fnet = F1 - F2
= 5.43 x 10^-3 N to the left
(B) all forces lies in the plane of loop.
hence net torque will be zero.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.