A single turn current loop carrying a current of 4.0A, is in the shape of a righ

Question

A single turn current loop carrying a current of 4.0A, is in the shape of a right triangle with sides 50cm, 120 cm, 130 cm. The loop is in a uniform magnetic field of magnitude 75 mT whose direction is parallel to the current in the 130 cm side of the loop. a.) Find the magnetic force on each of the three sides of the loop. b.) show that the total magnetic force on the loop is zero. c.) Calculate the magnetic dipole moment. d.) Calculate the magnitude and direction of the torque acting on the current carrying loop.

Explanation / Answer

given that

  We establish coordinates such that thetwo sides of the right triangle meet at the origin, and theLy = 50cm side runs along the +y axis, while theLx= 120 cm side runs along the +x axis. The anglemade by the hypotenuse (of length 130 cm) is = tan–1(50/120) = 22.6°, relative to the120 cmside. If one measures the angle counterclockwisefrom the +x direction, then the angle for the hypotenuse is180° –22.6° = +157°.
    Since we are only asked to find themagnitudes of the forces, we have the freedom to assume the currentis flowing, say, counterclockwise in the triangular loop (as viewedby an observer on the +z axis. We take B to be in the samedirection as that of the current flow in the hypotenuse. Then, withB= |B|= 0.0750T

Bx= - BCos = -0.0692 T and By= B Sin= 0.0288 T.

(a) Eq. 28-26 produces zeroforce when L|| B so there is noforce exerted on the hypotenuse of length 130 cm.

(b) On the 50 cm side, the Bxcomponent produces a force i LyBx k , and there is nocontribution from the By component. Using SI units, the magnitudeof the force on the Ly side istherefore

(4.00A)(0.500m)(0.0692T) = 0.138N.

(c) On the 120 cm side, theBy component produces a force iLxByK, and there is no contribution from the Bxcomponent. Using SI units, the magnitude of the force on theLx side is also

(4.00A)(1.20m)(0.0288T) = 0.138N.

keeping in mind thatBx < 0 due to our initial assumptions. If we hadinstead assumed B went the opposite direction of the current flowin the hypotenuse, then B x > 0 but By< 0 and a zero net force would still be the result.

magnetic dipole moment = NIA

                                         =(1) (4.0A)(1/2) base x heigth

                                          = (1) ( 4.0A) ( 0.5)(0.5 m)(1.2m)

                                          =1.2 A.m^2

magneitude of torque = xB

                                    = (1.2 A.m^2)(75 x 10^-3 T) sin 90

                                   = 90 x 10^-3 N .m

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