Question

A solenoid 60 cm long and of radius 4.0 cm has 3 layers of windings of 300 turns each. A 2.0 cm long wire of mass 2.5 g lies inside solenoid (near its centre) normal to its axis: both the wire and axis of the solenoid are in the horizontal plane. The wire is connected through two leads parallel to the axis of the solenoid to an external battery which supplies a current of 6.0 A in the wire. What value of current (with appropriate sense of circulation) in the windings of the solenoid

Answer 1

108.3 A

Answer 2

To solve this problem, we need to balance the gravitational force acting on the wire with the magnetic force exerted on it by the solenoid's magnetic field.

1. Calculate the gravitational force on the wire ($F_g$): The mass of the wire ($m$) is 2.5 g = $2.5 \times 10^{-3}$ kg. The acceleration due to gravity ($g$) is 9.8 m/s². $F_g = m \cdot g$ $F_g = (2.5 \times 10^{-3} \text{ kg}) \cdot (9.8 \text{ m/s}^2)$ $F_g = 0.0245 \text{ N}$

2. Determine the magnetic field inside the solenoid ($B$): The length of the solenoid ($L$) is 60 cm = 0.60 m. The solenoid has 3 layers of windings, with 300 turns each. Total number of turns ($N$) = 3 layers $\times$ 300 turns/layer = 900 turns. The magnetic field inside a long solenoid is given by: $B = \mu_0 \frac{N}{L} I_{solenoid}$ where $\mu_0$ is the permeability of free space ($4\pi \times 10^{-7}$ T·m/A), and $I_{solenoid}$ is the current in the solenoid windings.

3. Calculate the magnetic force on the wire ($F_m$): The current in the wire ($I_{wire}$) is 6.0 A. The length of the wire ($l_{wire}$) is 2.0 cm = 0.02 m. The wire lies normal to the solenoid's axis, meaning the angle ($\theta$) between the current direction in the wire and the magnetic field is 90°. The magnetic force on a current-carrying wire in a magnetic field is given by: $F_m = I_{wire} \cdot l_{wire} \cdot B \cdot \sin\theta$ Since $\theta = 90^\circ$, $\sin\theta = 1$. $F_m = I_{wire} \cdot l_{wire} \cdot B$

4. Equate the forces for equilibrium: For the wire to be suspended, the upward magnetic force must balance the downward gravitational force: $F_m = F_g$ $I_{wire} \cdot l_{wire} \cdot B = m \cdot g$

5. Substitute $B$ and solve for $I_{solenoid}$: Substitute the expression for $B$ into the equilibrium equation: $I_{wire} \cdot l_{wire} \cdot \left(\mu_0 \frac{N}{L} I_{solenoid}\right) = m \cdot g$

Now, solve for $I_{solenoid}$: $I_{solenoid} = \frac{m \cdot g \cdot L}{I_{wire} \cdot l_{wire} \cdot \mu_0 \cdot N}$

Plug in the values: $I_{solenoid} = \frac{(2.5 \times 10^{-3} \text{ kg}) \cdot (9.8 \text{ m/s}^2) \cdot (0.60 \text{ m})}{(6.0 \text{ A}) \cdot (0.02 \text{ m}) \cdot (4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}) \cdot (900)}$

Calculate the numerator: Numerator = $2.5 \times 10^{-3} \times 9.8 \times 0.60 = 0.0147$

Calculate the denominator: Denominator = $6.0 \times 0.02 \times 4\pi \times 10^{-7} \times 900$ Denominator = $0.12 \times 4\pi \times 10^{-7} \times 900$ Denominator = $(0.12 \times 900) \times 4\pi \times 10^{-7}$ Denominator = $108 \times 4\pi \times 10^{-7}$ Denominator = $432\pi \times 10^{-7}$ Using $\pi \approx 3.14159$: Denominator $\approx 432 \times 3.14159 \times 10^{-7} \approx 1357.168 \times 10^{-7} = 1.357168 \times 10^{-4}$

$I_{solenoid} = \frac{0.0147}{1.357168 \times 10^{-4}}$ $I_{solenoid} \approx 108.31 \text{ A}$

6. Determine the sense of circulation: The gravitational force acts downwards. For the wire to be suspended, the magnetic force must act upwards. The wire and the solenoid axis are in the horizontal plane. The wire is normal to the axis. Let's assume the solenoid axis is along the x-axis, and the wire is along the y-axis. Gravity is along the -z direction. We need the magnetic force to be along the +z direction. Using Fleming's Left-Hand Rule (or $\vec{F} = I(\vec{L} \times \vec{B})$): If the current in the wire ($I_{wire}$) is along the +y direction, then for the force ($\vec{F}$) to be along the +z direction, the magnetic field ($\vec{B}$) must be along the -x direction. To produce a magnetic field along the -x direction (using the right-hand rule for a solenoid), if you view the solenoid from the positive x-axis end, the current in the windings must circulate clockwise.

If the current in the wire ($I_{wire}$) is along the -y direction, then for the force ($\vec{F}$) to be along the +z direction, the magnetic field ($\vec{B}$) must be along the +x direction. To produce a magnetic field along the +x direction, if you view the solenoid from the positive x-axis end, the current in the windings must circulate counter-clockwise.

The problem does not specify the direction of current in the wire relative to the solenoid's ends. Therefore, the "appropriate sense of circulation" means that the current must circulate in a direction such that the magnetic field produced by the solenoid results in an upward force on the wire. For example, if the current in the wire flows from left to right across the solenoid's axis, and the solenoid's axis extends horizontally away from you, then the magnetic field must be directed towards you for the force to be upwards. This would require a specific circulation (e.g., clockwise when viewed from the end closer to you).

The value of the current required in the windings is approximately 108.3 A.