Moving Charges And Magnetism

repel each other

The magnetic field at the centre of a current carrying loop is given by

The magnetic moment at the centre of current carrying loop is given by
Thus, (given)
When both the current and the radius are doubled, the ratio becomes

So, the correct option is (A) : .

The magnetic field at the centre of a current carrying loop is given by

The magnetic moment at the centre of current carrying loop is given by
Thus, (given)
When both the current and the radius are doubled, the ratio becomes

So, the correct option is (A) : .

This question requires us to understand the magnetic field distribution in a toroid with thick windings. Let's analyze the problem step-by-step.

A toroid is a coil shaped like a donut, and its magnetic field properties are unique. The inner radius of the toroid is , and the outer radius is . The number of turns is , and it carries a steady current .

The magnetic field inside the toroid varies with the radial distance from the center. We use Ampère's Law to deduce this:

In simpler terms, inside a toroid, the magnetic field is directly proportional to current and inversely proportional to the radial distance , particularly between the inner radius and the outer radius . The magnetic field is zero outside this range ( or ) since the windings do not exist there to enclose the magnetic field.

Therefore, the relationship of the magnetic field with respect to radial distance is as follows:

• For ,
• For ,
• For ,

The graph that correctly represents this variation will show zero field from the center to , an inversely proportional decrease between and , and again zero beyond .

Therefore, the correct graph should exhibit a region of zero magnetic field for , a declining section between and , and again zero beyond . The last image matches this description accurately.

Thus, the correct variation of the magnetic field with radial distance is shown by the image above.

1. Magnetic Field inside a Long Solenoid:

The magnetic field inside a tightly wound long solenoid with ‘n’ turns per unit length carrying a current I is given by:

where is the permeability of free space, is the number of turns per unit length, and is the current.

2. Force on a Charged Particle in a Magnetic Field:

A particle with charge ‘q’ moving with velocity in a magnetic field experiences a magnetic force given by the Lorentz force formula:

Since the particle is projected perpendicular to the axis of the solenoid (which is the direction of the magnetic field), the velocity is perpendicular to . Therefore, the magnitude of the force is:

3. Circular Motion of the Charged Particle:

The magnetic force acts perpendicular to the velocity of the particle, causing it to move in a circular path in a plane perpendicular to the magnetic field. The magnetic force provides the necessary centripetal force for this circular motion.

Centripetal force , where is the mass of the particle and is the radius of the circular path.

Equating the magnetic force to the centripetal force:

4. Radius of the Circular Path:

From the above equation, we can solve for the radius of the circular path:

5. Condition for the Particle Not to Strike the Solenoid:

For the particle not to strike the solenoid, the radius of the circular path must be less than or equal to the radius of the solenoid ‘r’. For the maximum speed, we consider the case where the radius of the circular path is exactly equal to the radius of the solenoid:

6. Substitute the Magnetic Field and Solve for Maximum Speed:

Substitute the expression for the magnetic field inside the solenoid, , into the equation for the radius:

Now, solve for the maximum speed :

7. Compare with the Given Options:

Comparing the derived expression for with the given options, we find that it matches option 1:

Final Answer: The final answer is

Step 1: Understanding the scenario
Initially, the particle moves horizontally with constant velocity due to the velocity selector. The electric field ( ) is vertically downward, and the magnetic field ( ) is into the plane of the paper. At time , the electric field is switched off, leaving only the magnetic field acting on the particle.

Step 2: Radius of the circular path after switching off the electric field
After the electric field is switched off, the particle moves under the influence of only the magnetic field, thus describing a circular path. The radius ( ) of this path is given by:

Step 3: Identify particle position at
The time period of circular motion under a magnetic field is given by:

Thus, the given time is exactly half of the time period ( ). At half the time period, the particle will have rotated exactly 180° from its original direction.

This means at , the particle will be horizontally opposite to the original position, now traveling towards the left.

Step 4: Position of the particle at the given time
Initially, particle was moving horizontally rightward along the line:

This indicates the particle was initially located at a vertical distance of from the origin (since , thus ).
Thus, the particle started at the topmost point of a circle of radius .

After half rotation (180°), the particle will be exactly at the bottom point of the circle, located at the same vertical distance below the initial line (thus, on the horizontal axis, ). The particle is now located exactly on the x-axis, passing through the origin.

At this moment, the particle’s velocity vector is horizontal to the left.

Step 5: Calculating angular momentum about origin O at this moment
Angular momentum ( ) about any point is given by:

Here, the position vector passes through the origin (since particle is exactly on the x-axis at the bottom of the circle), making the particle's position vector and velocity vector exactly collinear (position vector along x-axis and velocity along negative x-axis).

The cross product of two collinear vectors is always zero:

Thus, the angular momentum about the origin O at this moment is zero.

Magnetic susceptibility of most materials tends to decrease as temperature increases. This relationship is typically governed by the Curie-Weiss law, which approximates susceptibility at a different temperature using the formula:
where:
- is the susceptibility at temperature ,
- is the susceptibility at a reference temperature ,
- is the new temperature.

Given and , , the susceptibility at 200 K is: