Askiitians Tutor Team
Last Activity: 5 Months ago
To tackle your question, we need to break it down into two parts: the work done in rotating the small coil and the ratio of distances related to the forces on the wires. Let's start with the first part regarding the coils.
Work Done in Rotating the Small Coil
When a small circular coil is placed at the center of a larger coil, both carrying the same current, they interact magnetically. The magnetic field produced by the larger coil affects the small coil. The work done in rotating the small coil about its diameter can be derived from the torque exerted on it due to the magnetic field.
Magnetic Field Calculation
The magnetic field \( B \) at the center of the larger coil can be calculated using the formula:
- B = \frac{{\mu_0 N I}}{{2R}}
Here, \( \mu_0 \) is the permeability of free space, \( N \) is the number of turns in the larger coil, \( I \) is the current, and \( R \) is the radius of the larger coil.
Torque on the Small Coil
The torque \( \tau \) acting on the small coil due to the magnetic field is given by:
- \tau = n \cdot B \cdot A \cdot I \cdot \sin(\theta)
Where \( n \) is the number of turns in the small coil, \( A \) is the area of the small coil (\( A = \pi r^2 \)), and \( \theta \) is the angle of rotation. When the coil is rotated by an angle \( \pi \), \( \sin(\theta) \) becomes 1.
Work Done Calculation
The work done \( W \) in rotating the coil through an angle \( \theta \) is given by:
- W = \int_0^{\theta} \tau \, d\theta
Substituting the expression for torque, we find:
- W = n \cdot B \cdot A \cdot I \cdot \theta
For \( \theta = \pi \), the work done becomes:
- W = n \cdot \frac{{\mu_0 N I}}{{2R}} \cdot \pi r^2 \cdot I \cdot \pi
This simplifies to:
- W = \frac{{n \mu_0 N I^2 \pi^2 r^2}}{{2R}}
Force on the Wires in Parallel
Now, let's move on to the second part of your question regarding the three wires connected in parallel with resistances in the ratio of 3:4:5. If the net force on the middle wire is zero, we need to find the ratio of distances \( d_1/d_2 \).
Understanding Forces on Wires
When wires carry current, they experience a magnetic force due to the magnetic field created by the other wires. The force \( F \) on a wire carrying current \( I \) in a magnetic field \( B \) is given by:
- F = I \cdot L \cdot B \cdot \sin(\phi)
Where \( L \) is the length of the wire and \( \phi \) is the angle between the current direction and the magnetic field. In this case, we assume \( \phi = 90^\circ \), so \( \sin(\phi) = 1 \).
Setting Up the Ratios
Let’s denote the resistances of the three wires as \( R_1 = 3x \), \( R_2 = 4x \), and \( R_3 = 5x \). The currents through these wires, based on the voltage \( V \), can be expressed as:
- I_1 = \frac{V}{3x}, I_2 = \frac{V}{4x}, I_3 = \frac{V}{5x}
The forces on the wires due to the magnetic fields created by the other wires can be expressed in terms of their respective currents. If the net force on the middle wire (wire 2) is zero, the forces exerted by wires 1 and 3 must balance each other.
Finding the Distance Ratio
Assuming the distances from the middle wire to the other two wires are \( d_1 \) and \( d_2 \), we can set up the equation:
Substituting the expressions for force, we get:
- \frac{V}{3x} \cdot d_1 = \frac{V}{5x} \cdot d_2
After canceling common terms, this simplifies to:
- \frac{d_1}{d_2} = \frac{5}{3}
Thus, the ratio of distances \( d_1/d_2 \) is \( 5:3 \).
In summary, the work done in rotating the small coil is given by a specific formula involving the magnetic field and the parameters of the coils, while the ratio of distances in the parallel wire scenario can be derived from the balance of forces acting on the wires. If you have any further questions or need clarification, feel free to ask!
