Question

A straight wire AB of mass 40 g and length 50 cm is suspended by a pair of flexible leads in uniform magnetic field of magnitude 0.40 T as shown in the figure. The magnitude of the current required in the wire to remove the tension in the supporting leads is (Take g = 10 ms^-2).

Answer 1

Correct Answer: 2

For equilibrium, the magnetic force on the wire must balance its weight: \[ F_{\text{magnetic}} = F_{\text{gravity}}. \] The magnetic force is given by: \[ F_{\text{magnetic}} = I L B, \] where \( I \) is the current, \( L = 50 \, \text{cm} = 0.5 \, \text{m} \) is the length of the wire, and \( B = 0.40 \, \text{T} \) is the magnetic field. The gravitational force is: \[ F_{\text{gravity}} = mg, \] where \( m = 40 \, \text{g} = 0.040 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \). Equating the forces: \[ I L B = mg. \] Solve for \( I \): \[ I = \frac{mg}{L B}. \] Substitute the values: \[ I = \frac{0.040 \cdot 10}{0.5 \cdot 0.40}. \] Simplify: \[ I = \frac{0.40}{0.2} = 2 \, \text{A}. \] Thus, the required current in the wire is \( \boxed{2 \, \text{A}} \).