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# A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends

Question:

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.

Solution:

Length of the rod, l = 0.45 m

Mass suspended by the wires, m = 60 g = 60 × 10−3 kg

Acceleration due to gravity, g = 9.8 m/s2

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.,

$B I l=m g$

$\therefore B=\frac{m g}{I l}$

$=\frac{60 \times 10^{-3} \times 9.8}{5 \times 0.45}=0.26 \mathrm{~T}$

A horizontal magnetic field of 0.26 T normal to the length of the conductor should be set up in order to get zero tension in the wire. The magnetic field should be such that Fleming’s left hand rule gives an upward magnetic force.

(b) If the direction of the current is revered, then the force due to magnetic field and the weight of the wire acts in a vertically downward direction.

∴Total tension in the wire = BIl + mg

$=0.26 \times 5 \times 0.45+\left(60 \times 10^{-3}\right) \times 9.8$

$=1.176 \mathrm{~N}$