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. There exists a uniform horizontal magnetic field B which is parallel to the longer side of the loop. The value of tension which is least is

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

Answer 1

The correct option is C $\frac{mg-2iBb}{2}$

Let us name the rectangle as PQRS as shown


And force due to magnetic field on PQ and RS is zero as they are parallel to Magnetic filed.

And force equations on other two sides is,

$T_1+ibB+T_2-ibB-mg=0$. ......(i)

$T_1+T_2=mg$

Torque about $B$ (shown in figure) gives us,

$ibBl+T_{1}l-mg\frac{l}{2}=0$

$T_1=\frac{mg}{2}-ibB$. ..........(ii)

And from (i) and (ii)

$T_2=\frac{mg}{2}+ibB$

Therefore minimum tension is $\frac{mg-2iBb}{2}$