Answer to Question #159813 in Electric Circuits for zain ul abdeen

Question

 The needle of a magnetic compass has magnetic moment 9.70 mA

m2. At its location, the Earth’s magnetic field is 55.0 mT north at 48.0° below the horizontal. (a) Identify the orientations of the compass needle that represent minimum potential energy and maximum potential energy of the needle–field system. (b) How much work must be done on the needle to move it from the former to the latter orientation? 

Answer 1

Given: $\mu = 9.70 \space mA*m^2 = 9.7*10^{-3} \space A* m^2$

$B = 55.0 \space \mu T = 55*10^{-6}\space T$

North at $48.0^{o}$ below horizontal

(a) orientation of needle for $U_{min}$ to $U_{max}$

(b) W = ? from $U_{min}$ to $U_{max}$

$U= -\overrightarrow{\mu}\overrightarrow{B} = - \mu B cos\theta$

For $U_{min}$, $\theta = 0^o$ it is parallel to direction of $\overrightarrow{B}$

pointing northward at $48^o$ below horizontal

For $U_{max},$ $\theta = 180^o$ pointing opposite of $\overrightarrow{B}$

south at $48^o$ above horizontal

$U_{min} = -\mu B cos0^o = -9.7*10^{-3} Am^2 * 5 5*10^{-6} cos0^o= -5.34*10^{-7}J$

$U_{max} = -\mu B cos180^o = -9.7*10^{-3} Am^2 * 5 5*10^{-6}cos180^o = 5.34*10^{-7}J$

(b) Conservation of Energy: $U_{min}+W = U_{max}$

$W = 5.34*10^{-7}-(-5.34*10^{-7}) = 1.07*10^{-6}J$