Magnetic Field

To solve this question, we need to determine the earth's resultant magnetic field at a place where the vertical component is given, and the angle of dip is provided.

Concept: The earth’s magnetic field at any place can be resolved into two components:

• Vertical component
• Horizontal component

The angle of dip ( ) is related to these components by the following relations:

Given:

• The vertical component
• Angle of dip

Let's calculate the earth's resultant magnetic field:

First, use the relation for the vertical component:

But we know:

Calculating :

therefore:

Now substitute the values:

Solving for :

Therefore, the earth's resultant magnetic field at that place is 1 × 10^-4 T.

Conclusion: The correct answer is:

1 × 10^-4 T

Magnetic force
⊥
⇒ W_b = 0
⇒ Δ_KE = 0 and speed remains constant.
The correct option is (A): Both (A) and (R) are true and (R) is the correct explanation of (A).

= in the material
=µ₀ without material

So, fractional increase is

So, the correct option is (A):

= =

x= ≃1.4

So, the correct option is (C): 1.4

To find the fractional increase in the magnetic field inside a solenoid when filled with a magnetic material, we need to understand the effect of magnetic susceptibility on the magnetic field inside the solenoid.

1. **Magnetic Field Inside a Solenoid:**

For a solenoid with air inside, the magnetic field is given by:

where:

• = permeability of free space
• = number of turns per unit length of the solenoid
• = current through the solenoid

2. **Effect of Magnetic Material:**

When a magnetic material with susceptibility is inserted inside the solenoid, the permeability becomes:

The new magnetic field inside the solenoid is:

3. **Fractional Increase in the Magnetic Field:**

The fractional increase in the magnetic field is given by:

Given , the fractional increase in the magnetic field is simply this susceptibility value:

Therefore, the correct answer is the fractional increase is .

So, the correct answer is (C) :

Step 1: Determine the Magnetic Field Due to the Semicircular Arc

The magnetic field at the center of a circular arc of radius R carrying current I is given by:

where is the permeability of free space, and is the angle subtended by the arc at the center in radians. For a semicircular arc, .

Step 2: Substitute the Given Values

Given , we have:

Step 3: Consider the Magnetic Field Due to the Straight Wires

The straight portions of the wire do not contribute to the magnetic field at point O. This is because the magnetic field lines due to an infinitely long straight wire form concentric circles around the wire. At the center of the semicircle, the straight sections are radial, so the magnetic field produced by them at point O will be perpendicular to the plane of the semicircle and thus won’t affect the field in the plane caused by the semicircular section.

Conclusion: The magnitude of the magnetic field at the center O is (Option 3).

The correct answer is 30.

Applying Kirchhoff's Voltage Law (KVL):

The differential equation for the circuit is given by:

The equation for the potential difference across the circuit is:

Substituting Known Values:

Substitute , , , and :

Simplify the equation:

Final Answer:

The correct option is (B):

T

T

T

Step 1: Write the Force Equation

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

Where:

• : Current
• : Length of the conductor in the magnetic field
• : Magnetic field strength

Step 2: Substitute in Terms of and

The current can be expressed as:

Substitute into the force equation:

Simplify:

Step 3: Substitute Known Values

Given:

Substitute these values into the equation:

Simplify:

Final Answer:

The magnetic force is

Step 1: Formula for Magnetic Intensity

The magnetic intensity inside a solenoid is given by:

Where:

• : Number of turns (60)
• : Current in Amperes
• : Length of the solenoid (15 cm = 0.15 m)

Step 2: Rearrange to Solve for

Rearrange the formula to solve for :

Step 3: Substitute the Values

Substitute , , and :

Simplify the expression:

Final Answer:

The current required to demagnetize the bar magnet is

Given:

• Number of turns ( ) = 60
• Length of the solenoid ( ) = 15 cm = 0.15 m
• Magnetic field ( ) =
• Permeability of free space ( ) =

Step 1: Formula for Magnetic Field in a Solenoid

The magnetic field inside a solenoid is given by the formula:

Where:

• : Magnetic field
• : Permeability of free space
• : Number of turns per unit length
• : Current in the solenoid

Step 2: Substituting Values

First, calculate the number of turns per unit length: Now substitute into the formula:

Step 3: Solve for Current ( )

Rearranging the formula for : Substituting the values: Simplify:

Step 4: Final Answer

The current in the solenoid is .

Statement I is correct: For diamagnetic substances, the magnetic susceptibility lies between , as diamagnetic substances are repelled by the magnetic field, and their susceptibility is negative.
Statement II is also correct: Diamagnetic substances move from stronger magnetic fields to weaker magnetic fields due to their negative susceptibility. This is because diamagnetic materials tend to create an opposing magnetic field that repels the external field, thus moving to areas where the magnetic field is weaker.
Therefore, both statements are correct.

When a bar magnet is dropped through a metallic cylindrical pipe, the phenomenon can be explained using Lenz's law and the induction of eddy currents:

• The changing magnetic flux due to the falling magnet induces eddy currents in the metallic pipe.
• According to Lenz’s law, these eddy currents generate their own magnetic field, which opposes the change in magnetic flux caused by the falling magnet.
• This opposing magnetic field creates an upward force on the magnet, reducing its acceleration and slowing its descent.

Comparison with a Non-Magnetic Bar:

A non-magnetic bar falling through the same pipe would not induce eddy currents because it does not have a magnetic field. The forces acting on the non-magnetic bar would be:

• Gravity, pulling it downward.
• Air resistance, which slightly opposes its motion.

Since there are no additional opposing forces (like those due to eddy currents), the non-magnetic bar falls more quickly than the magnet.

Analysis of Assertion and Reason:

• Assertion (A): The bar magnet takes longer to fall through the pipe than a non-magnetic bar of the same geometry and mass.
• Reason (R): Eddy currents in the metallic pipe create opposing magnetic fields, which slow down the descent of the magnet.

Both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).

Final Answer:

Both Assertion (A) and Reason (R) are true, and Reason (R) correctly explains Assertion (A).

Step 1: Analyzing the given statements
Statement I: Diamagnetic properties are independent of temperature. Therefore, Statement I is incorrect.
Statement II: In diamagnetic materials, the induced magnetic dipole moment is opposite to the external magnetic field, which makes Statement II true.

Thus, the correct answer is that Statement I is incorrect, and Statement II is true.

Solution:
Given:

T

(plane perpendicular to the field)

The area of the loop is .
The magnetic flux is . Induced emf: $ \mu V$.

Answer: 50.265

We are given the length of a rod and its angular speed. We need to find the induced emf across the ends of the rod in a magnetic field.

Solution

1. Given Values:

Length of the rod,

Angular speed,

Magnetic field, (assumed from the calculation)

2. Convert rpm to Radians per Second:

Using the approximate value of π as 22/7:

3. Use the Formula for Induced EMF:

The induced emf is given by:

4. Substitute the Known Values:

Final Answer

Thus, the induced emf is 88 mV.

The problem asks for the work done in rotating a current-carrying coil in a uniform magnetic field from an initial orientation to a final one. We are given the coil's properties, the current it carries, the strength of the magnetic field, and the angle of rotation.

Concept Used:

The work done in rotating a magnetic dipole (such as a current-carrying coil) in a uniform magnetic field is equal to the change in its potential energy. The potential energy of a magnetic dipole in a magnetic field is given by the dot product of the magnetic moment and the magnetic field:

where is the magnitude of the magnetic dipole moment, is the magnitude of the magnetic field, and is the angle between the magnetic moment vector and the magnetic field vector .

The magnetic moment of a coil with turns, area , and carrying current is:

The work done in changing the orientation of the coil from an initial angle to a final angle is:

Step-by-Step Solution:

Step 1: Determine the initial and final angles ( and ).

The magnetic moment vector is always perpendicular to the plane of the coil. The initial state is given as "the plane of the coil is perpendicular to the magnetic field". This means the magnetic moment vector is parallel to the magnetic field vector . Therefore, the initial angle is:

The coil is then turned through . The final angle is:

Step 2: Calculate the magnitude of the magnetic dipole moment ( ).

We are given:

• Number of turns,
• Area,
• Current,

Using the formula :

Step 3: Calculate the work done using the formula.

The work done is . We have:

• Magnetic field,

Substituting these values:

Step 4: Convert the work done to microjoules ( ).

Since , we can convert the result:

The work done in turning the coil is 100 .

Step 1: Magnetic Field Due to the Arc

For the arc with radius and angle , the magnetic field at the origin is:

Step 2: Magnetic Field Due to the Straight Segment

For the straight segment of the wire:

Step 3: Magnetic Field Due to Other Segments

Since the magnetic field due to the straight segments at the origin is zero:

Step 4: Calculate the Total Magnetic Field

Thus, the total magnetic field at the origin is:

Given: an electron (mass , charge ) enters a uniform magnetic field with initial velocity (where ). The de-Broglie wavelength initially is . We need to find its wavelength after time .

Concept Used:

The magnetic field exerts a force on the moving electron given by the Lorentz force:

This force changes the direction of velocity but not its magnitude (since the magnetic force is perpendicular to ). Hence, the speed remains constant.

The de-Broglie wavelength is given by:

Since (the magnitude of velocity) is unchanged, the de-Broglie wavelength also remains constant in time.

Step-by-Step Solution:

Step 1: Write the equation of motion:

The magnetic field causes circular motion with angular frequency (cyclotron frequency):

Step 2: The magnitude of velocity remains , only its direction changes with time. Therefore, momentum magnitude remains constant.

Step 3: Hence, the de-Broglie wavelength at time is:

Final Computation & Result:

The de-Broglie wavelength of the electron remains unchanged with time.