A conducting bar PQ of length l carrying curr | vedclass

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(C) $1$. The forces acting on the conducting bar $PQ$ are:$(a)$ The gravitational force $Mg$ acting downwards.$(b)$ The magnetic force $F_m = I(l 	im

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$(Mg - IBl)/2$

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(C) $1$. The forces acting on the conducting bar $PQ$ are:$(a)$ The gravitational force $Mg$ acting downwards.$(b)$ The magnetic force $F_m = I(l 	imes B)$ acting on the bar. According to Fleming's left-hand rule,with current $I$ flowing from $P$ to $Q$ and the magnetic field $B$ directed into the plane,the magnetic force acts vertically upwards.$(c)$ The tension $T$ in each of the two strings acting vertically upwards.$2$. For the bar to be in equilibrium,the net force acting on it must be zero.$3$. Summing the vertical forces: $2T + F_m - Mg = 0$.$4$. Substituting $F_m = IBl$,we get $2T + IBl - Mg = 0$.$5$. Rearranging for $T$,we have $2T = Mg - IBl$,which gives $T = (Mg - IBl)/2$.

$A$ conducting bar $PQ$ of length $l$ carrying current $I$ is suspended from a rigid support as shown in the figure. $A$ uniform magnetic field $B$ perpendicular to $PQ$ and directed away from the reader (into the plane) is applied. If the mass of the bar is $M$,find the tension in each string.

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