Question 11 Mark
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.
- What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?
- 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.
Answer
View full question & answer→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 = 5A
- Magnetic field (B) is equal and opposite to the weight of the wire i.e.,
BIl = mg
$\therefore\text{B}=\frac{\text{mg}}{\text{Il}}$
$=\frac{60\times10^{-3}\times9.8}{5\times0.45}=0.26\ \text{T}$
- A horizontal magnetic field of 0.26 T normal to the length of the conductor should be set,
$=0.26\times5\times0.45+(60\times10^{-3})\times9.8$
$=1.176\ \text{N}$






