Question

A square loop wire of length l= 0.1m on each side had mass 50 g and pivots about axis AA' that corresponds to a horizontal side of the square. A magnetic field of 500 G directed vertically downward, uniformly fills the region in the vicinity of the loop. The loop carries current I so that it is in equilibrium at angle 20 degree. Consider force on each side separately and find the direction of current to maintain 20 degree? Find the torque in the loop about the axis?

Answer 1

Answer:-

(i)

$\vec{F}=I\vec{l}\times\vec{B}$

$|\vec{F}_{AB}|=IlBsin(90+\theta)=IlBcos(\theta)$

$|\vec{F}_{BC}|=IlBsin(\theta)$

$|\vec{F}_{CD}|=IlBsin(90-\theta)=IlBcos(\theta)\\ |\vec{F}_{AD}|=IlBsin(\theta)$

The current in the loop should be counter-clockwise in

order to maintain balance.

(ii) $\tau=\frac{l}{2}|\vec{F_{AD}}| + l|\vec{F_{CD}}|-\frac{l}{2}|\vec{F_{BC}}|=lIlB(\frac{sin\theta}{2}+cos\theta -\frac{sin\theta}{2})$

$\therefore \ \ \boxed{\tau=Il^2Bcos\theta}$