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 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.) $g = 9.8\; m s^{-2}.$
Length of the rod, $l=0.45\, m$
Mass suspended by the wires, $m=60\,\, g=60 \times 10^{-3}\, kg$
Acceleration due to gravity, $g=9.8 \,m / s ^{-2}$
Current in the rod flowing through the wire, $I=5\, A$
$(a)$ Magnetic field $(B)$ is equal and opposite to the weight of the wire i.e.,
$B I l=m g$
$\therefore B=\frac{m g}{I l}$
$=\frac{60 \times 10^{-3} \times 9.8}{5 \times 0.45}=0.26\, T$
A horizontal magnetic field of $0.26\, T$ normal to the length of the conductor should be set up.
$(b)$ When the direction of the current is reversed, $BIl$ and mg will act downwards. So the effective tension in the wires is
$T=0.26 \times 5 \times 0.45+\left(60 \times 10^{-3}\right) \times 9.8$
$=1.176\, N$
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