Magnetic Force

To solve this problem, we need to determine the value of current for two long, parallel wires that are 0.20 m apart and carry the same current in the same direction. The force of attraction per meter between the wires is given as .

The formula for the force per unit length between two parallel current-carrying wires is given by:

where:

• , is the permeability of free space.
• and are the currents through the wires, both are equal to A.
• , is the distance between the wires.

Since the currents are the same and in the same direction, we can write:

This simplifies to:

Setting this equal to the given force per unit length:

Simplifying gives:

Solving for , we get:

Rounding to a single decimal place, the approximate value of is . Thus, the correct answer is 1.4.

The potential drop per unit length (potential gradient) along the potentiometer wire is calculated using the formula:
ℇ = V/L
where V is the voltage across the potentiometer wire and L is its length.

The total resistance in the circuit is the resistance of the potentiometer wire plus the resistance R from the resistance box:
Total resistance = 20 Ω + R

The current i flowing through the circuit is given by:
i = ℰ / (20 Ω + R)
with ℰ = 4 V (emf of the standard cell).

The potential drop across the potentiometer wire:
V = i × 20 Ω

Therefore, the potential gradient along the wire is:
Potential gradient = V / 300 cm = (4 V × 20 Ω) / ((20 Ω + R) × 300 cm)

At the null point (60 cm), the potential difference is equal to the emf of the 20 mV cell:
(Potential gradient) × 60 cm = 20 mV

Substitute the potential gradient equation:
((4 × 20) / (20 + R) × 300) × 60 = 0.02

Simplifying:
(80 / (20 + R)) × (60 / 300) = 0.02
(80 / (20 + R)) × (1/5) = 0.02
80 / (20 + R) = 0.1

Cross-multiplying gives:
80 = 0.1 × (20 + R)
800 = 20 + R
R = 780 Ω

Since 780 Ω is within the given range (780,780), the solution is validated.

The force on a straight conductor in a magnetic field is given by:

Here:

• (since is perpendicular to the 5 cm side).

Substitute the values:

The force is given as . Equating:

Thus, the value of is 9.

Given:

Step 1: Recall the Magnetic Force Formula

The magnetic force ( ) experienced by a charged particle with charge ( ) moving with velocity ( ) in a magnetic field ( ) is given by:

Step 2: Analyze the Force on the Electron

Inside a solenoid, the magnetic field ( ) is uniform and directed along the axis of the solenoid.

• The electron is moving along the axis of the solenoid, so its velocity ( ) is parallel to the magnetic field ( ).
• The cross product ( ) is zero when is parallel to .

Thus, the magnetic force on the electron is:

Step 3: Determine the Motion of the Electron

Since the magnetic force on the electron is zero, there is no net force acting perpendicular to its motion. As a result:

• The electron will continue to move along the axis of the solenoid with constant velocity.
• It will not be accelerated, nor will it follow a curved or parabolic path.

Final Answer:

The electron will not experience a magnetic force (B), and it will continue to move along the axis of the solenoid (C).

This problem requires us to calculate the magnetic field at the center of a square current-carrying loop and then determine the value of a constant from the given expression for the magnetic field.

Concept Used:

The magnetic field at a point P, due to a straight wire of finite length carrying a current , at a perpendicular distance from the wire is given by the Biot-Savart law application:

where and are the angles subtended by the ends of the wire at point P. For a square loop, the total magnetic field at the center is the sum of the magnetic fields due to its four sides. By symmetry, the field from each side is identical in magnitude and direction.

Step-by-Step Solution:

Step 1: Identify the given parameters.

Current in the loop, .

Side length of the square loop, .

The loop is in air, so we use the permeability of free space, .

Step 2: Determine the geometry for calculating the field from one side.

The perpendicular distance from the center of the square to the midpoint of any side is half the side length.

The angles and subtended by the ends of a side at the center are equal. The diagonals of a square bisect at , so the lines from the center to the vertices of a side make angles of with the perpendicular from the center to that side.

Step 3: Calculate the magnetic field due to one side of the square loop.

Using the formula for a finite wire:

Substitute the values:

Step 4: Calculate the total magnetic field at the center of the loop.

The total magnetic field is four times the field from a single side.

To match the given format, we can write this as:

Step 5: Compare the calculated magnetic field with the given expression to find X.

The problem states that the magnetic field at the center is .

By comparing both sides of the equation, we find:

The value of X is 40.