Solenoids and Toroids MCQ Quiz - Objective Question with Answer for Solenoids and Toroids - Download Free PDF

Last updated on May 20, 2025

Latest Solenoids and Toroids MCQ Objective Questions

Solenoids and Toroids Question 1:

The magnetic field inside a 200 turns solenoid of radius 10 cm is 2.9 × 10–4 Tesla. If the solenoid carries a current of 0.29 A, then the length of the solenoid is __________ π cm.

Answer (Detailed Solution Below) 8

Solenoids and Toroids Question 1 Detailed Solution

Explanation:

Assuming long solenoid

\(\mathrm{B}=\mu_{0}\left(\frac{\mathrm{~N}}{\ell}\right) \mathrm{i} \)

\(\ell=\frac{\mu_{0} \mathrm{Ni}}{\mathrm{~B}}=\frac{\left(4 \pi \times 10^{-7}\right)(200)(0.29)}{2.9 \times 10^{-4}} \mathrm{~m} \)

∴ Lenght of the solenoid = 8π cm

Solenoids and Toroids Question 2:

A solenoid of length 1.5 m and 4 cm diameter possesses 10 turns per cm. A current of 5A is flowing through it, the magnetic induction at axis inside the solenoid is

(μ0 = 4π × 10-7 weber amp-1 m-1)

  1. 4π × 10-7 gauss
  2. 2π × 10-7 gauss
  3. 4π × 10-5 tesla
  4. 2π × 10-5 tesla

Answer (Detailed Solution Below)

Option 4 : 2π × 10-5 tesla

Solenoids and Toroids Question 2 Detailed Solution

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

B = μ₀ × n × I

Where:

  • μ₀ = 4π × 10-7 weber amp-1 m-1 (permeability of free space)
  • n = number of turns per unit length = 10 turns per cm = 10 × 102 turns/m
  • I = current passing through the solenoid = 5A

Now, substituting the values into the equation:

B = (4π × 10-7) × (10 × 102) × 5

B = 2π × 10-5 tesla

The magnetic induction at the axis inside the solenoid is 2π × 10-5 tesla.

Solenoids and Toroids Question 3:

The magnetic field inside a 200 turns solenoid of radius 10 cm is 2.9 × 10–4 Tesla. If the solenoid carries a current of 0.29 A, then the length of the solenoid is __________ π cm.

Answer (Detailed Solution Below) 8

Solenoids and Toroids Question 3 Detailed Solution

Explanation:

Assuming long solenoid

\(\mathrm{B}=\mu_{0}\left(\frac{\mathrm{~N}}{\ell}\right) \mathrm{i} \)

\(\ell=\frac{\mu_{0} \mathrm{Ni}}{\mathrm{~B}}=\frac{\left(4 \pi \times 10^{-7}\right)(200)(0.29)}{2.9 \times 10^{-4}} \mathrm{~m} \)

∴ Lenght of the solenoid = 8π cm

Solenoids and Toroids Question 4:

A tightly wound long solenoid carries a current of 1.5 A. An electron is executing uniform circular motion inside the solenoid with a time period of 75ns. The number of turns per metre in the solenoid is_______.

[Take mass of electron me = 9 × 10–31 kg, charge of electron |qe| = 1.6 × 10–19 C, 

\(\left.\mu_{0}=4 \pi \times 10^{-7} \frac{\mathrm{~N}}{\mathrm{~A}^{2}}, 1 \mathrm{~ns}=10^{-9} \mathrm{~s}\right]\)

Answer (Detailed Solution Below) 250

Solenoids and Toroids Question 4 Detailed Solution

Calculation:

Since period of a revolving charge is \(\frac{2 \pi \mathrm{~m}}{\mathrm{qB}}\)

Where B = magnetic field

due to a solenoid = µ0 nI 

∴ \(\mathrm{T}=\frac{2 \pi \mathrm{~m}}{\mathrm{q}\left(\mu_{0} \mathrm{nI}\right)}\)

\(75 \times 10^{-9}=\frac{(2 \pi)\left(9 \times 10^{-31}\right)}{1.6 \times 10^{-19} \times 4 \pi \times 10^{-7} \times \mathrm{n} \times 1.5}\)

∴ N = 250 

Solenoids and Toroids Question 5:

A long solenoid of diameter 0.1 m has 2 × 104 turns per meter. At the centre of the solenoid a coil of 100 turns and radius 0.01 m is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to 0 A from 4 A in 0.05 s. If the resistance of the coil is 10π2 Ω, then the total charge flowing through the coil during this time is ___________.

Fill in the blank with the correct answer from the options given below.

  1. 16 μC
  2. 32 μC
  3. 16 π μC
  4. 32 π μC

Answer (Detailed Solution Below)

Option 4 : 32 π μC

Solenoids and Toroids Question 5 Detailed Solution

Given:

Number of turns, n = 100

Radius, r = 0.01 m

Resistance, R = 10π² Ω

Concept and Explanation:

From Faraday's Law, induced emf is:

ε = −N (dϕ / dt)

Since ε = IR and I = Δq / Δt, we get:

ε / R = −N / R × (dϕ / dt) ⇒ ΔI = −N / R × (dϕ / dt)

So, Δq = −[N / R × (Δϕ / Δt)] × Δt

The negative sign indicates that the induced emf opposes the change in magnetic flux (Lenz’s Law).

Substitute into the final formula:

Δq = (μ₀ × n × π × r²) / R

Putting values:

Δq = [4π × 10⁻⁷ × 100 × 4 × π × (0.01)²] / (10π²)

Δq = 32 μC

Top Solenoids and Toroids MCQ Objective Questions

Magnetic Field inside a solenoid is ________.

  1. increases from one end to another
  2. uniform
  3. varies from point to point
  4. None of the above

Answer (Detailed Solution Below)

Option 2 : uniform

Solenoids and Toroids Question 6 Detailed Solution

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CONCEPT:

  • 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.

F1 P.Y Madhu 14.04.20 D 3

  • 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:-

\(B=\frac{{{\mu }_{0}}NI}{l}\)

Where, N = number of turns, = length of the solenoid,  l = current in the solenoid and μo = absolute permeability of air or vacuum.

EXPLANATION:

  • The magnetic field inside a solenoid is uniform. So option 2 is correct.

Two solenoids having lengths L and 2L and the number of loops N and 4N, both have the same current, then the ratio of the magnetic field will be

  1. 1 : 2
  2. 2 : 1
  3. 1 : 4
  4. 4 : 1

Answer (Detailed Solution Below)

Option 1 : 1 : 2

Solenoids and Toroids Question 7 Detailed Solution

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CONCEPT:

  • solenoid is an instrument that consists of copper coiling over a cylinder designed to create a strong magnetic field inside the coil because of the flow of current through the coil.
  • By wrapping the same wire many times around a cylinder, the magnetic field due to the flow of current can become quite strong.
  • Hence we can say that the strength of the magnetic field will change as a current through coil or number of turn’s changes.
  • Magnetic field strength is independent of the diameter of the cylinder of a solenoid.
  • The strength of the magnetic field in a solenoid will be directly proportional to the number of turns and amount of current flowing through a wire and will be inversely proportional to its length.
  • Thus it is given by \(B = \frac{{{μ _0}NI}}{l}\)

Where N = number of turns and I = current, l = length of the solenoid

CALCULATION:
Given - Length of 1st solenoid = L, Length of 2nd solenoid = 2L, number of turns of 1st solenoid = N and number of turns of 2nd solenoid = 4L
  • Magnetic field due to the solenoid is given by
\(\Rightarrow B = \frac{{{μ _0}NI}}{l}\)
  • As μo and current (I) is constant, therefore 

\(\Rightarrow \;B \propto \frac{N}{L}\;\)

\( \Rightarrow \;\frac{{{B_1}}}{{{B_2}}}\; = \;\frac{{{N_1}}}{{{N_2}}} \times \frac{{{L_2}}}{{{L_1}}}\; = \;\frac{N}{{4N}} \times \frac{{2l}}{L}\; = \;\frac{1}{2}\)

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= ___________.

  1. μonI2
  2. nI/μo
  3. μonI
  4. nI2o

Answer (Detailed Solution Below)

Option 3 : μonI

Solenoids and Toroids Question 8 Detailed Solution

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Concept:

F1 P.Y Madhu 14.04.20 D 3

  • 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 -

\(B=\frac{{{\mu }_{0}}NI}{l}\)

Where,

N = number of turns, 

= length of the solenoid,  

l = current in the solenoid

μo = absolute permeability of air or vacuum.

Explanation:

From the above explanation, we can see that the magnetic field inside the solenoid is 

\(B=\frac{{{\mu }_{0}}NI}{l}\)

And for unit length  (l = 1) the above equation can be modified as

\(B={{{\mu }_{0}}NI}\)

Hence option 3 is correct among all

What happens to the strength of electromagnet if the soft iron core is put into it?

  1. It will increase
  2. It will decrease
  3. It will remain constant
  4. Initially, it will increase and then decrease.

Answer (Detailed Solution Below)

Option 1 : It will increase

Solenoids and Toroids Question 9 Detailed Solution

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CONCEPT:

  • An electromagnet is a type of magnet in which the magnetic field is produced by an electric current.
  • The magnetic field disappears when the current is turned off.

EXPLANATION:

  • When a soft iron core is inserted inside the solenoid then the strength of the magnetic field becomes very large because the iron core gets magnetized by induction.
  • The soft iron core helps in concentrating the magnetic lines of forces through the solenoid so that the magnetic field is almost uniform at the end face of the core.
  • Hence, the strength of the magnetic field increases when a soft iron core is inserted inside a solenoid. Therefore, option 1 is correct.

Railways Solution Improvement Satya 10 June Madhu(Dia)

  • The strength of an electromagnet can be increased by increasing the number of turns of wire around the iron core and by increasing the current or voltage.

The magnetic field inside a long solenoid:

  1. is zero
  2. is uniform
  3. increases from the axis towards the surface
  4. decreases from the axis towards the surface

Answer (Detailed Solution Below)

Option 2 : is uniform

Solenoids and Toroids Question 10 Detailed Solution

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CONCEPT:

  • solenoid is an instrument that consists of copper coiling over a cylinder designed to create a strong magnetic field inside the coil because of the flow of current through the coil.
    • By wrapping the same wire many times around a cylinder, the magnetic field due to the flow of current can become quite strong.
    • Hence we can say that the strength of the magnetic field will change as a current through coil or number of turns changes.
    • Magnetic field strength is independent of the diameter of the cylinder of a solenoid.
  • The strength of the magnetic field in a solenoid will be directly proportional to the number of turns and amount of current flowing through a wire and will be inversely proportional to its length and it is given as

\(\Rightarrow B = \frac{{{μ _0}NI}}{l}\)

Where N = number of turns and I = current, l = length of the solenoid

EXPLANATION:

Magnetism ALP Rishi 12Q Part 1 Hindi - Final images Q10

  • Inside a long current carrying straight solenoid, magnetic field lines are straight and parallel.
  • The strength of parallel magnetic field lines indicates that the magnetic field is uniform everywhere inside a long solenoid.​ Therefore option 2 is correct.

A solenoid has length L, radius r and N turns of wire. If a current i is passed through this solenoid, the magnetic field inside the solenoid is independent of ______

  1. length L
  2. radius r
  3. turns of wire N
  4. current i

Answer (Detailed Solution Below)

Option 2 : radius r

Solenoids and Toroids Question 11 Detailed Solution

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CONCEPT:

F1 J.K Madhu 10.07.20 D3

  • Solenoid: A type of electromagnet that generates a controlled magnetic field through a coil wound into a tightly packed helix.
    • The uniform magnetic field is produced when an electric current is passed through it.
  • The magnetic field inside a solenoid is proportional to the applied current and the number of turns per unit length.
  • The magnetic field inside a solenoid does not depend on the radius of the solenoid.
  • The field inside is constant.

B = μ0 N i

where N is no. of turns per unit length, i is current in the solenoid and μ0 is the permeability of free space.

EXPLANATION:

The field inside a solenoid is given by:

B = μ0 N i

  • Here N is the number of turns per unit length. So change in length will change the value of N.
  • So The magnetic field inside a solenoid depends on the applied current and the number of turns, and the length of solenoid.
  • The magnetic field inside a solenoid does not depend on the radius of it.
  • Hence the correct answer is option 2.

The magnetic field inside a long current-carrying solenoid

  1. decreases as we move towards the ends of the solenoid. 
  2. increases as we move towards the ends of the solenoid. 
  3. is zero along the axis of the solenoid. 
  4. is the same at all points inside the solenoid. 

Answer (Detailed Solution Below)

Option 4 : is the same at all points inside the solenoid. 

Solenoids and Toroids Question 12 Detailed Solution

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Concept:

Solenoid:

  • The solenoid is a type of electromagnet, the purpose of which is to generate a controlled magnetic field through a coil wound into a tightly packed helix.
  • The magnetic field is formed around the coil when an electric current passes through it and draws the plunger in.

  • The magnetic field at the center of the solenoid, B = μ0 nI, where, n = number of turns per unit length, I = current 

Explanation:

  • The magnetic field inside a long straight solenoid-carrying current is the same at all points.
  • It is because the magnetic field in the solenoid is constant because the lines are completely parallel to each other.

By inserting a soft iron piece into the solenoid, the strength of the magnetic field

  1. decrease
  2. increase
  3. first increase then decrease
  4. remains unchanged

Answer (Detailed Solution Below)

Option 2 : increase

Solenoids and Toroids Question 13 Detailed Solution

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Concept:

  • Electromagnets are the kind of magnet where the wire is wrapped around an iron core
  • A solenoid is a device that produces a uniform magnetic field inside it.

  • The magnetic field at any point beyond the solenoid is very thin.
  • The magnetic field inside the solenoid is parallel to its axis at all times.
  • The magnetic field produced by a solenoid can be defined using the Law of the Ampere, B = μ0nI, where, n = number of turns per unit length, I = current flowing

  • When a soft iron piece is introduced inside the solenoid, this iron core is magnetized thereby increasing the magnetic field inside a solenoid.
  • Therefore, a soft iron core inserted inside a solenoid increases the strength of the magnetic field.

Hence, the correct option is 2.

An electric bulb is connected in series with a solenoid and they are connected with an AC supply. When a soft iron rod is inserted in solenoid then the intensity of electric bulb will be

  1. unchanged
  2. decrease
  3. increase
  4. fluctuate

Answer (Detailed Solution Below)

Option 2 : decrease

Solenoids and Toroids Question 14 Detailed Solution

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CONCEPT:

  • An electromagnet is a type of magnet in which the magnetic field is produced by an electric current.
  • The magnetic field disappears when the current is turned off.

EXPLANATION:

  • When a soft iron core is inserted inside the solenoid then the strength of the magnetic field becomes very large because the iron core gets magnetized by induction.
  • The soft iron core helps in concentrating the magnetic lines of forces through the solenoid so that the magnetic field is almost uniform at the end face of the core.
  • Hence, the strength of the magnetic field increases when a soft iron core is inserted inside a solenoid.
  • This increases the inductance of the solenoid. So, the inductive reactance of the solenoid increases.
  • Consequently, a large fraction of the applied AC voltage appears across the solenoid. As a result of this, there is less voltage across the bulb and the brightness of the bulb decreases. Therefore, option 2 is correct.

Railways Solution Improvement Satya 10 June Madhu(Dia)

  • The strength of an electromagnet can be increased by increasing the number of turns of wire around the iron core and by increasing the current or voltage.

The strength of magnetic field inside a long current carrying straight solenoid is

  1. more at the ends than at the centre
  2. minimum in the middle
  3. same at every points
  4. found to increase from one end to the other

Answer (Detailed Solution Below)

Option 3 : same at every points

Solenoids and Toroids Question 15 Detailed Solution

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The correct answer is same at every point.

Key Points

Concept:

  • solenoid is an instrument that consists of copper coiling over a cylinder designed to create a strong magnetic field inside the coil because of the flow of current through the coil.
    • By wrapping the same wire many times around a cylinder, the magnetic field due to the flow of current can become quite strong.
    • Hence we can say that the strength of the magnetic field will change as a current through coil or number of turns changes.
    • Magnetic field strength is independent of the diameter of the cylinder of a solenoid.
  • The strength of the magnetic field in a solenoid will be directly proportional to the number of turns and amount of current flowing through a wire and will be inversely proportional to its length and it is given as

\(\Rightarrow B = \frac{{{μ _0}NI}}{l}\)

Where N = number of turns and I = current, l = length of the solenoid

Explanation:

Magnetism ALP Rishi 12Q Part 1 Hindi - Final images Q10

  • Inside a long current carrying a straight solenoid, magnetic field lines are straight and parallel.
  • The strength of parallel magnetic field lines indicates that the magnetic field is uniform everywhere inside a long solenoid.​
  • i.e. The strength of the magnetic field inside a long current-carrying solenoid is uniform everywhere.
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