Question

The wire loop with current I, is placed in a magnetic field B = B ((cosa)i + (cosß)+ (cosy)ť) as shown. (a) Find the magnetic force acting on each side of this loop. What is the net magnetic force acting on the loop? (6) Find the initial magnetic dipole moment for this loop of current. (C) Find the initial magnetic torque acting on this loop of current. (d) Find the change of magnetic potential energy of the loop as it aligns with this given magnetic field. [15] y (0,2,0) C IB A (L,0,0) 2

Answer 1

Given:

magnetic field = $\vec{B}=B(cos\alpha\hat{i}+cos\beta\hat{j}+cos\gamma\hat{k})$

The area vector of loop =

A = Lhk

Magnetic force on a current-carrying wire is given by [where L = length of the wire]

Fri( IB)

magnetic dipole moment = $\vec{M}=i\vec{A}$

Torque on a current-carrying loop is given by

= i( AB

Potential energy of a current-carrying loop is given by

M.B - AB

a) Magnetic force on the side OA = $\vec{F}_{OA}=i(\vec{L}_{OA}\times \vec{B}) =i(L\hat{i}\times B(cos\alpha\hat{i}+cos\beta\hat{j}+cos\gamma\hat{k}))$

$=iLB(-cos\gamma\hat{j}+cos\beta\hat{k})$ [answer]

Magnetic force on the side AB = $\vec{F}_{AB}=i(\vec{L}_{AB}\times \vec{B}) =i(h\hat{j}\times B(cos\alpha\hat{i}+cos\beta\hat{j}+cos\gamma\hat{k}))$

   $=ihB(cos\gamma\hat{i}-cos\alpha\hat{k})$ [answer]

Magnetic force on the side BC = $\vec{F}_{BC}=i(\vec{L}_{BC}\times \vec{B}) =i(-L\hat{i}\times B(cos\alpha\hat{i}+cos\beta\hat{j}+cos\gamma\hat{k}))$

$=iLB(cos\gamma\hat{j}-cos\beta\hat{k})$ [answer]

Magnetic force on the side CD = $\vec{F}_{CD}=i(\vec{L}_{CD}\times \vec{B}) =i(-h\hat{j}\times B(cos\alpha\hat{i}+cos\beta\hat{j}+cos\gamma\hat{k}))$

   $=ihB(-cos\gamma\hat{i}+cos\alpha\hat{k})$ [answer]

Net magnetic force on the loop = $\vec{F}_{net}=\vec{F}_{AB}+\vec{F}_{CD}+\vec{F}_{EF}+\vec{F}_{GH}$

= 0 [answer]

b) The magnetic dipole moment = $\vec{M}=i\vec{A} = ihL\hat{k}$ [answer]

c) Initial torque on the loop = $\vec{\tau}=i(hL\hat{k}\times B(cos\alpha\hat{i}+cos\beta\hat{j}+cos\gamma\hat{k}))$

$=ihLB(-cos\beta\hat{i}+cos\alpha \hat{j})$ [answer]

d) Initial potential energy = $U_{i}=-ihL\hat{k}.B(cos\alpha\hat{i}+cos\beta\hat{j}+cos\gamma\hat{k})=-ihLBcos\gamma$

The unit vector in the direction of B is $\hat{n}=\frac{(cos\alpha\hat{i}+cos\beta\hat{j}+cos\gamma\hat{k})}{\sqrt{cos^{2}\alpha+cos^{2}\beta+cos^{2}\gamma}}$

therefore, when the loop is aligned with the magnetic field, then new area vector is $\vec{A'}=hL\left (\frac{(cos\alpha\hat{i}+cos\beta\hat{j}+cos\gamma\hat{k})}{\sqrt{cos^{2}\alpha+cos^{2}\beta+cos^{2}\gamma}} \right )$

Final potential energy of the loop  $U_{f}=\left [-ihL\left (\frac{(cos\alpha\hat{i}+cos\beta\hat{j}+cos\gamma\hat{k})}{\sqrt{cos^{2}\alpha+cos^{2}\beta+cos^{2}\gamma}} \right ).B(cos\alpha\hat{i}+cos\beta\hat{j}+cos\gamma\hat{k}) \right ]$

  $=-ihLB\sqrt{cos^{2}\alpha+cos^{2}\beta+cos^{2}\gamma}$

Therefore, change in potential energy is $\Delta U = U_{f}-U_{i}=-ihLB\sqrt{cos^{2}\alpha+cos^{2}\beta+cos^{2}\gamma}-(-ihLBcos\gamma)$

  $=ihLB\left (cos\gamma-\sqrt{cos^{2}\alpha+cos^{2}\beta+cos^{2}\gamma} \right )$​​​​​​​ [answer]