Question

A straight wire of length \(30 \mathrm{~cm}\) and mass 60 milligram lies in a direction \(30^{\circ}\) east of north. The earth's magnetic field at this site is horizontal and has a magnitude of \(0.8 \mathrm{G}\). What current must be passed through the wire so that it may float in air ? \(\left[\mathrm{g}=10\left(\mathrm{~m} / \mathrm{s}^{2}\right)\right]\) (a) \(10 \mathrm{Amp}\) (b) \(20 \mathrm{Amp}\) (c) \(40 \mathrm{Amp}\) (d) \(50 \mathrm{Amp}\)

Short answer

The required current for the wire to float in air is approximately \(50 A\), which corresponds to option (d).

Step-by-step solution

Calculate the weight of the wire

First, we need to find the gravitational force acting on the wire. The weight of the wire can be determined using the formula: Weight (W) = mass (m) × gravitational acceleration (g) Given, mass of the wire (m) = 60 milligram = 60 × 10^(-3) kg (converting to kg) Gravitational acceleration (g) = 10 m/s² Weight (W) = 60 × 10^(-3) kg × 10 m/s² = 0.6 N

Set up the equilibrium condition

The wire floats in the air due to the balance between the magnetic force (F_m) and the weight of the wire (W). The magnetic force on the wire can be calculated using this formula: F_m = B × I × L × sin(θ) Where: - B is the magnetic field (0.8 G = 0.8 × 10^(-4) T, converting to Tesla) - I is the current that we need to find - L is the length of the wire (30 cm = 0.3 m, converting to meters) - θ is the angle between the wire and the direction of the magnetic field (30 degrees) Since the wire is floating, the magnetic force (F_m) is equal to the weight (W), so: F_m = W

Solve for the current (I)

Using the equilibrium condition (F_m = W), we can solve for the current (I): 0.8 × 10^(-4) T × I × 0.3 m × sin(30°) = 0.6 N Dividing both sides by (0.8 × 10^(-4) T × 0.3 m × sin(30°)), we get: I = 0.6 N / (0.8 × 10^(-4) T × 0.3 m × sin(30°)) I = 0.6 N / (0.00008 T × 0.3 m × 0.5) I ≈ 50 A So, the required current for the wire to float in air is approximately 50 A, which corresponds to option (d).