Question

A 2A current carrying straight metal wire of resistance 1 Ω, resistivity 2 × 10⁻⁶ Ωm, area of cross-section 10 mm² and mass 500 g is suspended horizontally in mid air by applying a uniform magnetic field B. The magnitude of B is ......... × 10⁻¹ T (given, g = 10 m/s²)

Answer 1

5

Answer 2

For the wire to be suspended in mid-air, the upward magnetic force must balance the downward gravitational force.

The magnetic force on a current-carrying wire of length $L$ in a uniform magnetic field $B$ is given by $F_B = BIL\sin\theta$. Since the wire is suspended horizontally, and the magnetic force must be vertically upwards to counteract gravity, the magnetic field must be perpendicular to the current (i.e., $\theta = 90^\circ$, so $\sin\theta = 1$).

Thus, $F_B = BIL$. The gravitational force is $F_g = mg$. Equating the forces: $BIL = mg$.

First, we need to find the length $L$ of the wire. We are given the resistance $R$, resistivity $\rho$, and area of cross-section $A$. The relationship between these is $R = \rho \frac{L}{A}$. From this, we can find $L$: $L = \frac{RA}{\rho}$

Given: Current $I = 2$ A Resistance $R = 1$ Ω Resistivity $\rho = 2 \times 10^{-6}$ Ωm Area of cross-section $A = 10$ mm² $= 10 \times 10^{-6}$ m² Mass $m = 500$ g $= 0.5$ kg Acceleration due to gravity $g = 10$ m/s²

1. Calculate the length of the wire (L): L = \frac{RA}{\rho} = \frac{(1 \text{ \(\Omega}) \times (10 \times 10^{-6} \text{ m}^2)}{2 \times 10^{-6} \text{ $\Omega$ m}} = \frac{10}{2} \text{ m} = 5 \text{ m})

2. Apply the force balance condition: $BIL = mg$ $B \times (2 \text{ A}) \times (5 \text{ m}) = (0.5 \text{ kg}) \times (10 \text{ m/s}^2)$ $10B = 5$ $B = \frac{5}{10} = 0.5 \text{ T}$

3. Express the magnitude of B in the required format: The question asks for $B$ in the format $X \times 10^{-1}$ T. $B = 0.5 \text{ T} = 5 \times 10^{-1} \text{ T}$

Thus, the magnitude of B is $5 \times 10^{-1}$ T.