$A$ straight horizontal conducting rod of length $0.45\; m$ and mass $60\; g$ is suspended by two vertical wires at its ends. $A$ current of $5.0\; A$ is set up in the rod through the wires.
$(a)$ What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?
$(b)$ What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field same as before? (Ignore the mass of the wires.) $g = 9.8\; m s^{-2}.$

(A) Given:
Length of the rod,$l = 0.45\; m$
Mass of the rod,$m = 60\; g = 60 \times 10^{-3}\; kg$
Acceleration due to gravity,$g = 9.8\; m s^{-2}$
Current in the rod,$I = 5.0\; A$
$(a)$ For the tension in the wires to be zero,the upward magnetic force must balance the downward weight of the rod.
$F_B = mg$
$BIl = mg$
$B = \frac{mg}{Il} = \frac{60 \times 10^{-3} \times 9.8}{5.0 \times 0.45} = 0.26\; T$
Thus,a magnetic field of $0.26\; T$ directed normal to the conductor is required.
$(b)$ When the current direction is reversed,the magnetic force $F_B$ acts downwards along with the weight $mg$.
Total tension $T = F_B + mg = BIl + mg$
$T = (0.26 \times 5.0 \times 0.45) + (60 \times 10^{-3} \times 9.8)$
$T = 0.588 + 0.588 = 1.176\; N$