MCQ Questions

Moving Charges and Magnetism Class 12 MCQs Questions with Answers

Class 12 Physics Chapter 4 MCQ Question 1.

Two charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field B = B₀k.

(A) They have equal z-components of momenta.
(B) They must have equal charges.
(C) They necessarily represent a particle- antiparticle pair.
(D) The charge to mass ratio satisfy:
\(\left(\frac{e}{m}\right)_{1}+\left(\frac{e}{m}\right)_{2}\) = 0
Answer:
(D) The charge to mass ratio satisfy:
\(\left(\frac{e}{m}\right)_{1}+\left(\frac{e}{m}\right)_{2}\) = 0

Explanation:
When charge/mass ratio of these two particles is same and charges on them are of opposite nature, then the charged particles will traverse identical helical paths in a completely opposite sense. Therefore, option (D) is correct.

Moving Charges And Magnetism MCQ Chapter 4 Question 2.

Biot-Savart law indicates that the moving electrons (velocity v) produce a magnetic field B such that

(A) B ⊥ n.
(B) B || u.
(C) it obeys inverse cube law.
(D) it is along the line joining the electron and point of observation.

Answer:
(A) B ⊥ n.

Explanation:
In Biot-Savart’s law, magnetic field B || idl x r and idl due to flow of electron is in opposite direction of v and by direction of cross product of two vectors, B.
dB = \(\frac{\text { I.dl } \sin \theta}{r^{2}}\)
or dB = \(\frac{I \times d l}{r}\)
According to Biot-Savart law, if magnetic field is not perpendicular to the motion of charge, then it will not move in helical path, which is not possible for motion of a charge in magnetic field. So, the magnetic field is perpendicular to the direction of flow of charge verifies answer ‘A’.

Moving Charges And Magnetism Class 12 MCQ Chapter 4 Question 3.

A current carrying circular loop of radius R is placed in the x – y plane with centre at the origin. Half of the loop with x > 0 is now bent so that it now lies in the y – z plane.

(A) The magnitude of magnetic moment now diminishes.
(B) The magnetic moment does not change.
(C) The magnitude of B at (O.O.z), z > >R increases.
(D) The magnitude of B at (O.O.z), z >>R is unchanged.
Answer:
(A) The magnitude of magnetic moment now diminishes.

Explanation:
For a circular loop of radius K, carrying current I in x – y plane, the magnetic moment M = 1 x πR₂ . It acts perpendicular to the loop along z – direction. When half of the current loop is bent in y – z plane, then magnetic moment due to half current loop is x – y plane, M₁ = I (πR₂/2) acting along x – direction Magnetic moment due to half current loop y – z plane, M₂ = I (πR₂) along x – direction. Net magnetic moment due to entire bent current loop,

M_net = \(\sqrt{\mathrm{M}_{1}^{2}+\mathrm{M}_{2}^{2}}\)
= \(\sqrt{2} \frac{I \pi R^{2}}{2}\)
= \(\frac{M}{\sqrt{2}}\)
Therefore, M_net < M or M diminishes.

Chapter 4 Physics Class 12 MCQ Question 4.

A circular current loop of magnetic moment M is in an arbitrary orientation in an external magnetic field B. The work done to rotate the loop by 30° about an axis perpendicular to its plane is

(A) MB
(B) \(\frac{\sqrt{3} \mathrm{MB}}{2}\)
(C) \(\frac{MB}{2}\)
(D) zero
Answer:
(D) zero

Explanation:
The work done to rotate the loop in magnetic field, W = MB (cos θ₁ – cos θ₂ When current carrying coil is rotated then there will be no change in angle between magnetic moment and magnetic field.
Here, θ₁ = θ₂ = a
⇒ W = MB (cos ∝ – cos ∝) = 0.

MCQ On Moving Charges And Magnetism Chapter 4 Question 5.

When a charge of 1C moving with velocity 1 m/s normal to a magnetic field experiences a force 1 N, then the magnitude of the magnetic field is

(A) 1 Gauss
(B) 1 Tesla
(C) 1 Orested
(D) None of the above
Answer:
(A) 1 Gauss

Explanation:
F = qvB sin θ
When q = 1C, υ = 1 m/s, F = 1N, θ = 90°, I then B = 1T

Moving Charges And Magnetism Class 12 MCQs Question 6.

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

(A) The electron will be accelerated along the axis.
(B) The electron path will be circular about the axis.
(C) The electron will experience a force at 45° to the axis and hence execute a helical path.
(D) The electron will continue to move with uniform velocity along the axis of the solenoid.
Answer:
(D) The electron will continue to move with uniform velocity along the axis of the solenoid.

Explanation:
The magnetic field inside the long current carrying solenoid is uniform. Therefore, magnitude of force on the electron of charge {- e) is given by F = – evB sin0 = 0 (0 = 0°) as magnetic field and velocity are parallel. The electron will continue to move with uniform velocity along the axis of the solenoid.

Ch 4 Physics Class 12 MCQ Question 7.

When a charged particle moves through a magnetic field perpendicular to its direction. Then

(A) Linear momentum changes
(B) kinetic energy remains constant
(C) Both (A) and (B)
(D) Both linear momentum and kinetic energy varies
Answer:
(B) kinetic energy remains constant

Explanation:
When a charged particle perpendicularly enters a magnetic field to the direction, the path of the motion is circular. In circular motion, the direction of velocity changes at every point (the magnitude remains constant). Therefore, the linear momentum changes at every point. But kinetic energy remains constant since the magnitude of velocity does not change.

Physics Class 12 Chapter 4 MCQ Question 8. 

length L of wire carries a steady current I. It is bent first to form a circular plane coil of one turn. A current flowing through it produces a magnetic field B at the centre of the coil. The same length is now bent more sharply to form a double loop of smaller radius. The magnetic field at the centre caused by the same current is

(A) B
(B) 2B
(C) 4B
(D) B/2
Answer:
(C) 4B

Explanation:
B₂ = n²B₁
Here n = 2, B₁ = B.
∴ B₂ = 4B

MCQ Of Chapter 4 Physics Class 12 Question 9.

A straight conductor carries a current from south to north. Point P and Q lie to the east and west of it at the same distance. The magnetic field at P is

(A) equal to magnetic field at Question
(B) smaller than the magnetic field at Question
(C) greater than the magnetic field at Question
(D) cannot be predicted unless the value if I is known.
Answer:
(A) equal to magnetic field at Question

Explanation:
B ∝ I, B ∝ 1/r
So, if I and r remains constant, then magnetic field at P = Magnetic field at Question

Class 12 Physics Ch 4 MCQ Question 10.

Magnetic field due to a straight solenoid at any point inside it is B = μ₀ni. Magnetic field at the end of the solenoid is

(A) B
(B) B/2
(C) 2B
(D) B/4
Answer:
(B) B/2

Explanation:
Magnetic field at the end of a current carrying solenoid is half of the magnetic field inside it.

MCQ Of Moving Charges And Magnetism Chapter 4 Question 11.

At any point, empty space surrounded by a toroid, the magnetic field is B₁. At any point, outside the toroid, the magnetic field is B₂.

(A) B₁> B₂
(B) B₂ > B₁
(C) B₁ = B₂
(D)B₁ = B₂ = 0
Answer:
(D)B₁ = B₂ = 0

Explanation:
As net current is zero, magnetic field at the empty space surrounded by toroid and outside the toroid is zero.

Physics Chapter 4 Class 12 MCQ Question 12.

An infinitely long straight conductor is bent into the shape as shown in the figure. Current in it is and the radius of the circular loop is r. The magnetic field at its centre is –

(A) Zero
(B) Infinite
(C) \(\frac{\mu_{0} i}{2 \pi r}(\pi-1)\)
(D) \(\frac{\mu_{0} i}{2 \pi r}(\pi+1)\)
Answer:
(C) \(\frac{\mu_{0} i}{2 \pi r}(\pi-1)\)

Explanation:
Magnetic field at O due to ABCD straight conductor = \(\frac{\mu_{0} i}{2 \pi r}\)
Magnetic field at O due to the BEC circular conductor = \(\frac{\mu_{0} i}{2 r}\)
The fields are in opposite direction. Hence the resultant field at O is \(\frac{\mu_{0} i}{2 r}-\frac{\mu_{0} i}{2 \pi r}=\frac{\mu_{0} i}{2 \pi r}\)(π – 1)

Class 12 Physics Chapter 4 MCQ Questions Question 13.

A solenoid of 1.5 metre length and 4.0 cm diameter has 10 turn per cm. A current of 5 A ampere is flowing through it. The magnetic field at axis inside the solenoid is

(A) 2K x 10^-3 T
(B) 2K x 10^-3 G
(C) 2n x 10^-7 T
(D) 2K x 10^-7 G
Answer:
(A) 2K x 10^-3 T

Explanation:
B = μ₀ni = 4π x 10^-7 x 5 x 10 x 10^-2 = 2π x 10^-3

Class 12 Chapter 4 Physics MCQ Question 14.

The strength of the magnetic field at distance r from a long straight current carrying wire is B. The field at a distance r/2 will be

(A) B
(B) 2B
(C) B/2
(D) B/4
Answer:
(B) 2B

Explanation: .
B ∝ 1/r
\(\mathrm{B}_{1} / \mathrm{B}_{2}=r_{2} / r_{1}=\frac{r / 2}{r}\)
∴B₂ = 2B₁ = 2B

MCQ Class 12 Physics Chapter 4 Question 15.

In a moving coil galvanometer, current in the coil is

(A) directly proportional to angle of deflection.
(B) inversely proportional to the angle of deflection.
(C) directly proportional to the square root of the angle of deflection.
(D) inversely proportional to the square root of the angle of deflection.
Answer:
(A) directly proportional to angle of deflection.

Explanation:
In a moving coil galvanometer, current in the coil is directly proportional to angle of deflection.

Chapter 4 MCQ Class 12 Physics Question 16.

Current sensitivity of a galvanometer is given by

(A) CG/nBA
(B) nBA/C
(C) nBA/CG
(D) CG/nBA
Answer:
(B) nBA/C

Explanation:
Current sensitivity of a galvanometer is the deflection produced when unit current passes through it.
Current sensitivity = θ/I = nBA/C

Moving Charges And Magnetism MCQs Chapter 4 Question 17.

The deflecting torque acting on the coil of a galvanometer is

(A) inversely proportional to number of turns.
(B) inversely proportional to current flowing.
(C) inversely proportional to area of the coil.
(D) directly proportional to the magnetic field strength.
Answer:
(D) directly proportional to the magnetic field strength.

Explanation:
τ = nBIA.
So, torque is directly proportional to the magnetic field strength, area of the coil, number of turns and current flowing.

Class 12th Physics Chapter 4 MCQ Question 18.

To convert a galvanometer to ammeter a shunt S is to be connected with the galvanometer. The effective resistance of the ammeter then is

(A) GS/(G + S)
(B) (G + S) / GS
(C) G + S
(D) None of the above
Answer:
(A) GS/(G + S)

Explanation:
Shunt (S) is connected in parallel to the galvanometer (resistance G). So, the effective resistance is GS/(G + S).

MCQ Questions For Class 12 Physics Chapter 4 Question 19.

A galvanometer of 100 Ω resistance gives full scale deflection for 10 mA current. To use it as an ammeter of 10 A range, the resistance of the shunt required is

(A) 10 Ω
(B) 0.10 Ω
(C) 0.01 Ω
(D) 0.001 Ω
Answer:
(B) 0.10 Ω

Explanation:
S = \(\frac{i_{g} G}{i-i_{g}}=\frac{100 \times 0.01}{10-0.01}\) = 0.1 Ω

Question 20.

An ammeter gives full scale deflection when current of 1.0 A is passed in it. It is converted into a 100 A range ammeter, What will be the ratio of the shunt resistance and its resistance ?

(A) 1 : 9
(B) 9 : 1
(C) 1 : 11
(D) 11 : 1
Answer:
(A) 1 : 9

Explanation:
S = \(\frac{i_{g} G}{i-i_{g}}\)
S/G = \(\frac{i_{g} G}{i-i_{g}}=\frac{1}{10-1}\) = 1.9

Question 21.

A galvanometer can be converted into a voltmeter by connecting a

(A) high resistance in series.
(B) high resistance in parallel.
(C) low resistance in parallel.
(D) low resistance in series.
Answer:
(A) high resistance in series.

Explanation:
To convert a galvanometer into a voltmeter, a high value resistance is to be connected in series with it.

Assertion And Reason Based MCQs (1 Mark each)

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false and R is true

Question 1.

Assertion (A): Magnetic field interacts with a moving charge only.
Reason (R): Moving charge produces a magnetic field.

Answer:
(A) Both A and R are true and R is the correct explanation of A

Explanation:
Current carrying wire creates magnetic field. This magnetic field has no effect on a stationary charge. But when the charge also moves, it creates a current. This current produces a magnetic field. Two fields interact and the charge is deflected. So, the assertion is true. Moving charge creates a current which produces a magnetic field. So, the reason is also true. Reason is the correct explanation of A.

Question 2.

Assertion (A): If an electron is not deflected when moving through a certain region of space, then the only possibility is that no magnetic field is present in that region.
Reason (R): Force on electron is directly proportional to the strength of the magnetic field.

Answer:
(A) Both A and R are true and R is the correct explanation of A

Explanation:
In absence of magnetic field, moving electron will not be deflected. This possibility is true. So, assertion is true.
\(\vec{F}=q(\vec{V} \times \vec{B})\). So, force on electron is directly proportional to the strength of the magnetic field. So, reason is true. Reason properly explains the assertion.

Question 3.

Assertion (A): The energy of a charged particle moving in a uniform magnetic field remains constant.
Reasoning (R): Work done by the magnetic field on the charge is zero.

Answer:
(A) Both A and R are true and R is the correct explanation of A

Explanation:
The force on a charged particle moving in a uniform magnetic field always acts in direction perpendicular to the direction of motion of the charge.
So work done by the magnetic field,
W = FS cos θ = FS cos 90° = 0
So, the energy of the charged particle does not change.
Both, assertion and reason are true and reason also explains the assertion.

Question 4.

Assertion (A): An electron and a proton moving with same velocity enters a magnetic field. The force experienced by the proton is more than the force experienced by the electron.
Reason (R): The mass of proton is more than the mass of the electron.

Answer:
(D) A is false and R is true

Explanation:
\(\vec{F}=q(\vec{V} \times \vec{B})\)
So, the force is mass independent. So, the assertion is false. Proton is obviously heavier than electron. So, reason is true. But reason does not explain the assertion.

Question 5.

Assertion (A): The magnetic field at the ends of a very long current carrying solenoid is half of that at the centre.
Reason (R): Magnetic field within a sufficiently long solenoid is uniform.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
Magnetic field inside a solenoid is B = μ₀ni.
Magnetic field at the end of a solenoid is 1/2 μ₀ni So, the assertion is true.
Magnetic field within a sufficiently long solenoid is uniform. So reason is also true. But it does not explain the assertion.

Question 6.

Assertion (A): The magnetic field produced by a current carrying solenoid is independent of its length and area of cross-section.
Reason (R): Magnetic field within a very long solenoid is uniform.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
Magnetic field inside solenoid B = p0m. It is independent of length and area of cross-section. Hence the assertion is true. Reason is also true. But it does not explain the assertion.

Question 7.

Assertion (A): A direct current flowing through a metallic rod produces magnetic field both inside and outside of the rod.
Reason (R): There is no flow of charge carrier inside the rod.

Answer:
(C) A is true but R is false

Explanation:
Charge carries flows through whole cross-section. So, the filed exists both inside and outside. So, the assertion is true and the reason is false.

Question 8.

Assertion (A): In moving coil galvanometer, the coil is wound on a metallic frame.
Reason (R): The metallic frame helps in making steady deflection without oscillation.

Answer:
(A) Both A and R are true and R is the correct explanation of A

Explanation:
Coil of a moving coil galvanometer is wound on a metal frame. So, the assertion is true. It is done to avoid any oscillation and fluctuating reading. The metal frame provides damping to reduce the escalation so that the reading becomes steady. So the reason is also true and properly explains the assertion.

Question 9.

Assertion (A): Torque on a coil is maximum when it is suspended radially in a magnetic field.
Reason (R): Torque tends to rotate a coil.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
The torque on the coil in a magnetic field is given by
τ = nlBA sin θ
For radial field, θ = 90° and sin θ = 1
Torque = nlBA and it is maximum.
So assertion is true.
Torque is the rotational equivalence of force. So, torque will tend to rotate a coil.
Reason is also true. But reason cannot explain the assertion that why the torque is maximum in the specified position.

Question 10.

Assertion (A): Galvanometer to ammeter conversion takes place by connecting a low value resistance in parallel with it. ‘
Reason (R): The low value resistance increases the effective resistance and protects the galvanometer.

Answer:
(C) A is true but R is false

Explanation:
Galvanometer to ammeter conversion takes place by connecting a low value resistance known as “shunt” in parallel with it. The assertion is true. When two resistors are connected in parallel then the effective resistance becomes lower than the lowest value of the two resistors. Hence the reason is false.

Question 11.

Assertion (A): Earth’s magnetic field does not affect the functioning of a moving coil galvanometer.
Reason (R): Earth’s magnetic field is too weak.

Answer:
(A) Both A and R are true and R is the correct explanation of A

Explanation:
The coil of moving coil galvanometer is suspended in a very strong radial magnetic field. Earth’s magnetic field is too weak compared to that and hence its effect is negligible. So, assertion and reason both are true and the reason explains the assertion properly.

Case-Based MCQs

Attempt any 4 sub-parts out of 5. Each sub-part carries 1 mark.

I. Read the following text and answer the following questions on the basis of the same:

Roget’s spiral:
Magnetic effects are generally smaller than electric effects. As a consequence, the force between currents is rather small, because of the smallness of the factor g. Hence, it is difficult to demonstrate attraction or repulsion between currents. Thus, for 5 A current in each wire at a separation of 1 cm, the force per metre would be 5 x 10^-4 N, which is about 50 mg weight. It would be like pulling a wire by a string going over a pulley to which a 50 mg weight is attached. The displacement of the wire would be quite unnoticeable.

With the use of a soft spring, we can increase the effective length of the parallel current and by using mercury, we can make the displacement of even a few mm observable very dramatically. You will also need a constant-current supply giving a constant current of about 5 A. Take a soft spring whose natural period of oscillations is about 0.5^-1 s. Hang it vertically and attach a pointed tip to its lower end, as shown in the figure here. Take some mercury in a dish and adjust the spring such that the tip is just above the mercury surface.

Take the DC current source, connect one of its terminals to the upper end of the spring and dip the other terminal in mercury. If the tip of the spring touches mercury, the circuit is completed through mercury. Let the DC source be put off to begin with. Let the tip be adjusted so that it just touches the mercury surface. Switch on the constant current supply and watch the fascinating outcome. The spring shrinks with a jerk, the tip comes out of mercury (just by a mm or so), the circuit is broken, the current stops, the spring relaxes and tries to come back to its original position, the tip again touches mercury establishing a current in the circuit and the cycle continues with tick, tick, tick,…

Question 1.

Magnetic effects:

(A) are equal to electric effects.
(B) are greater than electric effects.
(C) are smaller than electric effects.
(D) cannot be compared with electric effects.
Answer:
(C) are smaller than electric effects.

Explanation:
Magnetic effects are generally smaller than electric effects.

Question 2.

The force 10^-3 N,is equivalent to:

(A) 100 mg
(B) 100 g
(C)10g
(D) 10 mg
Answer:
(A) 100 mg

Explanation:
10^-3N = mass in kg x g in m/s² Or, 10^-3 = mass x 10
Mass = 10^-4 kg = 100 mg

Question 3.

Why the spring shrinks in Roget’s spiral ?

(A) The spring functions as a solenoid
(B) Due to force acting between two current carrying wires
(C) Due to magnetic effect of current
(D) Since the spring is soft.
Answer:
(B) Due to force acting between two current carrying wires

Explanation:
The spring shrinks due to force acting between two current carrying wires.

Question 4.

What are the main 3 components in a Roget’s spiral?

(A) Mercury, AC voltage source
(b) Mercury, DC voltage source
(C) Mercury, DC voltage source, key
(D) Mercury, AC voltage source, key
Answer:
(C) Mercury, DC voltage source, key

Explanation:
Mercury, DC voltage source, key is essential components for the Roget’s spiral to work.

Question 5.

What else can be used instead of mercury in Roget’s spiral ?

(A) Any liquid
(B) Water
(C) Kerosene oil
(D) Only mercury, nothing else
Answer:
(D) Only mercury, nothing else

Explanation:
Only mercury can be used in Roget’s spiral since mercury is a liquid metal through which an electrical circuit, may be completed.

II. Read the following text and answer the following questions on the basis of the same:
Galvanometer can sense/measure current. Improved mirror galvanometer was developed by William Thomson, later to become Lord Kelvin, in 1858. Thomson intended the instrument to read weak signal currents on very long submarine telegraph cables. The fundamental problems of transmitting/ receiving a signal through a lengthy submarine cable was that the electrical current tended to be very low (as little as 1/100,000th of a standard light bulb). So, it was very difficult to detect it.

To solve the problem it was thought that larger amount of electric current would be sent through the line. But Thomson had a different approach. He thought the best response was to devise a device that could read faint signals. The galvanometer, first invented in 1802, was a means of detecting electric current. It consisted of a needle that was deflected by the magnetic field created by the electric current. But the galvanometers of the day couldn’t detect the weak signals that came through a long underwater cable.

But the improved version of galvanometer was highly sensitive to detect the lowest current. The mirror galvanometer consists of a long fine coil of silk-covered copper wire. In the heart of that coil, within a little air-chamber, a small round mirror is hung by a single fibre of floss silk, with four tiny magnets cemented to its back.

A beam of light is thrown from a lamp upon the mirror, and reflected by it upon a white screen or scale a few feet distant, where it forms a bright spot of light; when there is no current on the instrument, the spot of light remains stationary at the zero position on the screen; but the instant a current traverses the long wire of the coil, the suspended magnets twist themselves horizontally out of their former position, the mirror is inclined with them, and the beam of light is deflected along the screen to one side or the other, according to the nature of the current.

If a positive electric current gives a deflection to the right of zero, a negative current will give a deflection to the left of zero, and vice versa. The air in the little chamber surrounding the mirror is compressed, so as to act like a cushion, and deaden the movements of the mirror; the mirror is thus prevented from idly swinging about at each deflections.

Question 1.

Improved mirror galvanometer was developed by

(A) Lord Kelvin
(B) Johann Schweigger
(C) Luigi Galvani
(D) Andre-Marie Ampere
Answer:
(A) Lord Kelvin

Explanation:
Improved mirror galvanometer was developed by William Thomson, later to become Lord Kelvin, in 1858.

Question 2.

Mirror galvanometer was primarily used to

(A) measure the current passing through electric bulb.
(B) measure the weak current received through lengthy submarine cable.
(C) measure current passing through human body.
(D) all of these.
Answer:
(B) measure the weak current received through lengthy submarine cable.

Explanation:
The fundamental problem was that the transmitting/receiving a signal through a lengthy submarine cable was very low. Instead of increasing the magnitude of the current transmission. Lord fcelvin modified the existing galvanometer s# that it became capable to measure the weakest current.

Question 3.

The basic principle of galvanometer is

(A) heating effect of current.
(B) torque developed by the electric current passing through a coil.
(C) magnetic effect of current.
(D) none of the above.
Answer:
(C) magnetic effect of current.

Explanation:
The galvanometer, was a means of detecting electric current. It consisted of a needle that was deflected by the magnetic field created by the electric current.

Question 4.

The mirror galvanometer consists of –

(A) a small round mirror attached to a fine coil of silk-covered copper wire.
(B) a long fine coil of silk-covered copper wire and a small round mirror hung by a single fibre of floss silk, with four tiny magnets cemented to its back.
(C) a small round mirror attached to four tiny magnets.
(D) None of the above
Answer:
(B) a long fine coil of silk-covered copper wire and a small round mirror hung by a single fibre of floss silk, with four tiny magnets cemented to its back.

Explanation:
The mirror galvanometer consists of a long fine coil of silk-covered copper wire. In the heart of that coil, within a little air- chamber, a small round mirror is hung by a single fibre of floss silk, with four tiny magnets cemented to its back

Question 5.

How the idly swinging of the mirror of mirror galvanometer is prevented?

(A) The little chamber surrounding the mirror was filled with a viscous liquid
(B) The mirror was placed in little chamber which was completely vacuum
(C) The mirror was attached to a spring
(D) The little chamber surrounding the mirror was filled with compressed air
Answer:
(D) The little chamber surrounding the mirror was filled with compressed air

Explanation:
The air in the little chamber surrounding the mirror is compressed, so as to act like a cushion, and deaden the movements of the mirror; the mirror is thus prevented from idly swinging about at each deflections.

III. Read the following text and answer the following questions on the basis of the same:

TOROID :
A toroid is a coil of insulated or enamelled wire wound on a donut-shaped form made of powdered iron. A toroid is used as an inductor in electronic circuits, especially at low frequencies where comparatively large inductances are necessary. A toroid has more inductance, for a given number of turns, than a solenoid with a core of the same material and similar size. This makes it possible to construct high-inductance coils of reasonable physical size and mass. Toroidal coils of a given inductance can carry more current than solenoidal coils of similar size, because larger-diameter wires can be used, and the total amount of wire is less, reducing the resistance.

In a toroid, all the magnetic flux is contained in the core material. This is because the core has no ends from which flux might leak off. The confinement of the flux prevents external magnetic fields from affecting the behaviour of the toroid, and also prevents the magnetic field in the toroid from affecting other components in a circuit. Standard toroidal transformers typically offer a 95% efficiency, while standard laminated transformers typically offer less than a 90% rating.

One of the most important differences between a toroidal transformer and a traditional laminated transformer is the absence of gaps. The leakage flux through the gaps contributes to the stray losses in the form of eddy currents (which is also expelled in the form of heat). A toroidal core doesn’t have an air gap. The core is tightly wound . The result is a stable, predictable toroidal core, free from discontinuities and holes.

Audible vibration or hum in transformers is caused by vibration of the windings and core layers from the forces between the coil turns and core laminations. The toroidal transformer’s construction helps quiet this noise. In audio, or signal transmitting applications, unwarranted noise will affect sound quality, so a transformer with low audible vibration is ideal. For this reason, many sound system engineers prefer to use a toroidal transformer instead of a traditional laminated transformer.

Question 1.

Toroid is a –

(A) fixed value resistor.
(B) capacitor.
(C) inductor.
(D) variable resistor.
Answer:
(C) inductor.

Explanation:
A toroid is a coil of insulated or enamelled wire wound on a donut-shaped form made of powdered iron. A toroid is used as an inductor in electronic circuits.

Question 2.

A toroid has inductance, for a given number of turns, than a solenoid with a core of same material and similar size.

(A) same
(B) more
(C) less
(D) variable
Answer:
(B) more

Explanation:
A toroid has more inductance, for a given number of turns, than a solenoid with a core of the same material and similar size. This makes it possible to construct high-inductance coils of reasonable physical size and mass.

Question 3.

Why inductance of solenoid is more than the inductance of a solenoid having same number of turns, core of same material and similar size?

(A) Core is endless hence there no leakage of flux.
(B) Resistance of wire is less hence magnitude of current flow is more
(C) Number of turns per unit length is more.
(D) Both (A) and (B)
Answer:
(A) Core is endless hence there no leakage of flux.

Explanation:
In a toroid, all the magnetic flux is contained in the core material. This is because the core has no ends from which flux might leak off.

Question 4.

Why sound system engineers prefer to use toroidal transformer?

(A) It is cheaper.
(B) It is lighter.
(C) It is compact.
(D) It does not create vibration or hum.
Answer:
(D) It does not create vibration or hum.

Explanation:
Audible vibration or hum in transformers is caused by vibration of the windings and core layers from the forces between the coil turns and core laminations. The toroidal transformer’s construction helps quiet this noise. For this reason, many sound system engineers prefer to use a toroidal transformer instead of a traditional laminated transformer.

Question 5.

Efficiency of toroidal transformer is around % which is than laminated core transformer.

(A) 95, lower
(B) 95, higher
(C) 50, lower
(D) 80, higher
Answer:
(B) 95, higher

Explanation:
Standard toroidal transformers typically offer a 95% efficiency, while standard laminated transformers typically offer less than a 90% rating.

MCQ Questions for Class 12 Physics with Answers