Question

A uniform conducting rectangular loop of sides $l,b$ and mass $m$ carrying current $i$ is hanging horizontally with the help of two vertical strings as shown in figure. There exists a uniform horizontal magnetic field $B$, which is parallel to the longer side of loop $l$. The least value of tension in strings is (Given acceleration due to gravity $=g$):

• A. $\frac{mg-iBb}{2}$
• B. $\frac{mg+iBb}{2}$
• C. $\frac{mg-2iBb}{2}$
• D. $\frac{mg+2iBb}{2}$

Answer 1

Answer: (C)

$\frac{mg-2iBb}{2}$

Let ${\overrightarrow{T}}_1$ and ${\overrightarrow{T}}_2$ be the tension in the wires when magnetic field $\overrightarrow{B}$ is applied.
If the rectangular loop is in equilibrium, $T_1+T_2\ne mg$, we need to take magnetic force into account.
Let us calculate torque about the point B. Net torque has to be zero.
${\tau }_{mg}=mg\frac{l}{2}$ anti-clockwise
${\tau }_{T_1}=T_{1}l$ clockwise
${\tau }_{F_m}=\overrightarrow{\mu }\times \overrightarrow{B}=iAB=ilbB$ clockwise
Equating both clockwise and anti-clockwise torques $T_{1}l+ilbB=mg\frac{l}{2}$
$\Rightarrow T_1=\frac{mg-2ibB}{2}$
Net force on the rectangular loop due to magnetic field is zero.
$\therefore T_1+T_2=mg$
$\Rightarrow T_2=mg-\left(\frac{mg-2ibB}{2}\right)=\frac{mg+2ibB}{2}$
Hence, least tension is $\frac{mg-2ibB}{2}$
Instead we can calculate $T_2$ by taking torque about point $A$.