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Test: Solenoid - NEET MCQ


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10 Questions MCQ Test Physics Class 12 - Test: Solenoid

Test: Solenoid for NEET 2024 is part of Physics Class 12 preparation. The Test: Solenoid questions and answers have been prepared according to the NEET exam syllabus.The Test: Solenoid MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Solenoid below.
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Test: Solenoid - Question 1

The force between two current carrying conductors is due to which of the following

Detailed Solution for Test: Solenoid - Question 1

we can say that any two current carrying conductors when placed near each other, will exert a magnetic force on each other.

Consider the system shown in the figure above.

Here, we have two parallel current carrying conductors, separated by a distance ‘d’, such that one of the conductors is carrying a current I1 and the other is carrying I2, as shown in the figure.

From the knowledge gained before, we can say that the conductor 2 experiences the same magnetic field at every point along its length due to the conductor 1.

The direction of magnetic force is indicated in the figure and is found using the right-hand thumb rule. The direction of the magnetic field, as we can see, is downwards due to the first conductor.

Test: Solenoid - Question 2

Which of the following laws give the direction of induced e.m.f

Detailed Solution for Test: Solenoid - Question 2

Lenz’s law is used for determining the direction of induced current.

Lenz’s law of electromagnetic induction states that the direction of induced current in a given magnetic field is such that it opposes the induced change by changing the magnetic field.

Following is the formula of Lenz’s law:

ϵ=−N (∂ϕB/∂t)

Where,

  • ε is the induced emf

  • ∂ΦB is the change in magnetic flux

  • N is the number of turns in the coil

Lenz’s law finds application in electromagnetic braking and in electric generators

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Test: Solenoid - Question 3

Magnetic Field inside a solenoid is ________.

Detailed Solution for Test: Solenoid - Question 3

Solenoid: A cylindrical coil of many tightly wound turns of insulated wire with a general diameter of the coil smaller than its length is called a solenoid.

  • A magnetic field is produced around and within the solenoid.
  • The magnetic field within the solenoid is uniform and parallel to the axis of the solenoid.

The strength of the magnetic field in a solenoid is given by:-

Where, N = number of turns, 
= length of the solenoid,  
l = current in the solenoid and
μo = absolute permeability of air or vacuum.
The magnetic field inside a solenoid is uniform. So option 2 is correct.

Test: Solenoid - Question 4

A solenoid coil of 300 turns /m is carrying a current of 5 A. Calculate the magnitude of magnetic intensity inside the solenoid.​

Detailed Solution for Test: Solenoid - Question 4

We know, Magnetic field of solenoid,
= μonI
=4πx10-7x300x5
=4πx15x10-5
=6πx10-9
=18.89x10-9
≈1.9x10-3T

Test: Solenoid - Question 5

There is a very long, very tightly wound solenoid in a round tube shape of small radius (compared to its length). It carries constant electric current. Now, it is bend to form a closed toroid. Consider the magnetic field inside the tube. It will  

Detailed Solution for Test: Solenoid - Question 5

Solenoid and Toroid:

  • solenoid is a coil of wire designed to create a uniform magnetic field inside when electric current flows through it.
  • The magnetic field inside the solenoid is given by the formula,
    B = μ0nI
    Where:
    • B is the magnetic field inside the solenoid,
    • 0) is the permeability of free space (4π×10−7T m/A),
    • n is the number of turns per unit length,
    • I is the current flowing through the solenoid.
  • toroid is a solenoid bent into the shape of a doughnut or a circular ring.
  • The magnetic field inside a tightly wound toroid is confined to the interior of the coil and is given by the formula,

    Where:
    • B is the magnetic field inside the toroid,
    • N is the total number of turns,
    • I is the current,
    • r is the average radius of the toroid.
  • The magnetic field remains confined within the tube and follows circular paths along the axis of the toroid.

Calculation:

Given,

Current, (I) = Constant electric current
Number of turns, (N)
Average radius of the toroid, (r)

The magnetic field inside the toroid is given by,

⇒ The magnetic field is inversely proportional to the radius (r). As (r) decreases, the magnetic field increases.

The magnetic field inside the toroid remains the same.

∴ The correct option is C.

Test: Solenoid - Question 6

An electron is projected with uniform velocity along the axis of a current carrying long solenoid. Which of the following is true?

Detailed Solution for Test: Solenoid - Question 6

→A solenoid consists of a helical winding of wire on a cylinder, usually circular in cross-section.
There can be hundreds or thousands of closely spaced turns, each of which can be regarded as a circular loop.
→Magnetic field due to solenoid B = μ0nl
μ0 = Permeability of free space
n = no. of windings per unit length
I = current.

→The direction of the field inside the solenoid is parallel to the axis, obtained by the right-hand thumb rule.
The direction of the field is shown in the figure.
→If a charged particle (q) is projected with uniform velocity along the axis of the solenoid then the force on the electron


Here, q = charge of the particle
 = velocity vector of the electron
 = magnetic field
θ = angle between 
→Now, when the particle (here electron) is projected along the axis of the solenoid the angle θ = 0° or 180° (as shown)


∴ F = qvB sin0 or qvB sin180
∴ F = 0
Hence, the electron will continue to move with uniform velocity.
So, the correct answer is option (b).

Test: Solenoid - Question 7

In a current carrying conductor varying electric field generates

Detailed Solution for Test: Solenoid - Question 7

According to Maxwell's equations, a varying electric field generates a magnetic field. This principle is fundamental to electromagnetism and is described by Faraday's Law of Induction and Ampère's Law with Maxwell's correction.

In a current-carrying conductor:

  1. The movement of charges (current) creates an electric field.
  2. A varying electric field generates a magnetic field around the conductor as described by Ampère’s Law.

Why other options are incorrect:

  1. B: Potential gradient:

    • A potential gradient is related to the electric field itself, not the generation of a magnetic field.
  2. C: Resistance:

    • Resistance is a property of the material and is not generated by a varying electric field.
  3. D: Current:

    • Current is the movement of charges due to an existing electric field, not a result of the varying electric field.
Test: Solenoid - Question 8

A long solenoid carrying a current produces a magnetic field B along its axis. If the current is doubled and the number of turns per cm is halved, the new value of magnetic field will be equal to

Detailed Solution for Test: Solenoid - Question 8

The magnetic field B inside a long solenoid is given by the formula:
⇒ B = μ0ni
Now i → 2i
And
n → n/2

∴ The new magnetic field remains the same as the original magnetic field B

Test: Solenoid - Question 9

An electron is projected with uniform velocity along the axis inside a current carrying long solenoid. Then :

Detailed Solution for Test: Solenoid - Question 9

       
Since the force (Lorentz force ∵ F = -e(V × B) on the electron due to the magnetic field is zero.
So it will move along the axis with uniform velocity.

Test: Solenoid - Question 10

Consider a long solenoid of 'n' turns per unit length and carrying a current 'I'. The magnetic field in the interior of the solenoid was shown to be given
by Bo= ___________.

Detailed Solution for Test: Solenoid - Question 10

  • A cylindrical coil of many tightly wound turns of insulated wire with generally diameter of the coil smaller than its length is called a solenoid.
  • A magnetic field is produced around and within the solenoid. The magnetic field within the solenoid is uniform and parallel to the axis of the solenoid.
  • Strength of the magnetic field in a solenoid is given by -


Where,
N = number of turns, 
= length of the solenoid,  
l = current in the solenoid
μo = absolute permeability of air or vacuum.
Now
we can see that the magnetic field inside the solenoid is 

And for unit length  (l = 1) the above equation can be modified as
B = μ0NI
Hence option C is correct among all

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