Magnetic Force on a Current-Carrying Wire MCQ Quiz - Objective Question with Answer for Magnetic Force on a Current-Carrying Wire - Download Free PDF

Calculation:

From motional emf,

e_max = BLV

∴ Heat produced = \(\frac{\mathrm{V}^2}{\mathrm{r}}=\frac{\mathrm{B}^2 \mathrm{~L}^2 \mathrm{~V}^2}{\mathrm{R}}\)

\(\mathrm{i}=\frac{\mathrm{BLV}}{\mathrm{R}}\)

|F| = BiL and P = F . V

∴ \(P=B\left(\frac{B L V}{R}\right) L V\)

= \(\frac{B^2 L^2 V^2}{R}\)

Concept Used:

To determine the ratio of the magnetic field induction at the origin due to the smaller coil to that of the bigger one, we use the formula for the magnetic field along the axis of a circular coil carrying current:

B = (μ₀ I a²) / (2 (x² + a²)^(3/2))

where:

B is the magnetic field at a distance x from the coil along its axis,

μ₀ is the permeability of free space,

I is the current in the coil,

a is the radius of the coil,

x is the distance from the coil to the point where the field is being calculated.

Calculation:

For the smaller coil (radius a and distance x), the magnetic field is:

B₁ = (μ₀ I a²) / (2 (x² + a²)^(3/2))

For the larger coil (radius 2a and distance 2x), the magnetic field is:

B₂ = (μ₀ I (2a)²) / (2 [(2x)² + (2a)²]^(3/2))

Simplifying:

B₂ = (μ₀ I 4a²) / (2 [4(x² + a²)]^(3/2)) = (μ₀ I 4a²) / (8(x² + a²)^(3/2)) = (μ₀ I a²) / (4(x² + a²)^(3/2))

The ratio of the magnetic field induction at the origin due to the smaller coil to that of the larger coil is:

Ratio = B₁ / B₂ = [(μ₀ I a²) / (2 (x² + a²)^(3/2))] / [(μ₀ I a²) / (4 (x² + a²)^(3/2))]

Simplifying:

Ratio = 4 / 2 = 2

Therefore, the ratio of the magnetic field induction at the origin due to the smaller coil to that of the bigger one is:

Option 1: 2 : 1

Calculation:

The maximum possible magnetic field is at the surface

\(\mathrm{B}_{\max }=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{a}} \)

\(\frac{\mathrm{~B}_{\max }}{2}=\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{a}} \)

It can be obtained inside as well as outside the wire

For inside, 

\( \frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{a}}=\frac{\mu_{0} \mathrm{Ir}}{2 \pi \mathrm{a}^{2}} \)

\(\Rightarrow \mathrm{r}=\frac{\mathrm{a}}{2} \)

For outside 

\(\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{a}}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}\)

⇒ r = 2a

∴ The correct answer is \(\left[\frac{\mathrm{a}}{2}, 2 \mathrm{a}\right]\)

The correct answer is Option 3: 2 A.

Calculation:

Given coercivity (H) = 140 A/m

Number of turns (N) = 112

Length of solenoid (L) = 1.6 m

Turns per unit length (n) = N/L = 112 / 1.6 = 70 turns/m

Using H = nI → 140 = 70 × I → I = 140/70 = 2 A

The correct answer is Option 1: Zero.

Explanation:

Magnetic Field (B): Generated around a current-carrying conductor.

Inside a thin conductor (idealized as infinitely thin): Magnetic field at the exact location of the conductor itself is considered zero.

At the surface of a thin wire, magnetic field induction is zero.

The magnetic field exists around the wire but not exactly at the conductor’s location.

CONCEPT:

• Fleming Left-hand rule gives the force experienced by a charged particle moving in a magnetic field or a current-carrying wire placed in a magnetic field.
  • This rule was originated by John Ambrose Fleming.
  • It is used in an electric motor.
  • It states that "stretch the thumb, the forefinger, and the central finger of the left hand so that they are mutually perpendicular to each other. If the forefinger points in the direction of the magnetic field, the central finger points in the direction of motion of charge, then the thumb points in the direction of force experienced by positively charged particles."

EXPLANATION:

• In Fleming's left rule, the middle finger represents the direction of current flowing through the conductor. So option 3 is correct.
• The thumb represents the direction of the magnetic force.
• The forefinger represents the direction of the magnetic field.

CONCEPT:

Fleming's Left-hand rule 

• It gives the force experienced by a charged particle moving in a magnetic field or a current-carrying wire placed in a magnetic field.
• It states that "stretch the thumb, the forefinger, and the central finger of the left hand so that they are mutually perpendicular to each other.
• If the forefinger points in the direction of the magnetic field, the central finger points in the direction of motion of charge, then the thumb points in the direction of force experienced by positively charged particles."

EXPLANATION:

According to question

1. Forefinger (Index finger): In the direction north 
2. Middle finger: In the direction east
3. The thumb is pointing out of the paper i.e., on top.

The force acting on the conductor will be out of the paper i.e., on top. Therefore option 3 is correct.

CONCEPT:

• Whenever a charge particle is moving in the magnetic field then there is a force acting on the charge particle, this force is called magnetic force.

The magnetic force on a charge particle of charge q and velocity V in a magnetic field B is given by:

\(\vec{F}=q~\left( \vec{V}\times \vec{B} \right)=q~V~B~Sin\theta\)

It is the cross product of velocity vector \(\vec{V}\) and magnetic field vector \(\vec{B}\) where θ is angle between velocity and magnetic field.

EXPLANATION:

• In the solenoid, the magnetic field is along the axis of the solenoid.
• We have projected the proton along the axis, so the angle between magnetic field and velocity is zero.

Here θ = 0°

\(Force~on~proton~\left( {\vec{F}} \right)=q~\left( \vec{V}\times \vec{B} \right)=q~V~B~Sin\theta =q~V~B~Sin0=0\)

CONCEPT:

• When a moving charged particle enters a magnetic field then the path followed by the charged particle is circular if the magnetic field is perpendicular to the velocity of the particle.
• More the radius of the path followed by the particle, the lesser will be the curvature, and the lesser the radius, the more will be curvature.
• To perform the circular motion, the required centripetal force would be provided by the magnetic force on the moving charge.
• The radius of the path followed by the charged particle moving in the magnetic field is given by:

\(\Rightarrow {\text{r}} = \frac{{{\text{mv}}}}{{{\text{Bq}}}}\)

where r = radius, m = mass, v = velocity, B = strength of the magnetic field, q = charge on the particle.

EXPLANATION:

• The radius of the path followed by the electron moving in the magnetic field is given by:

\(\Rightarrow {\text{r}} = \frac{{{\text{mv}}}}{{{\text{Be}}}}\)

The above equation can be written as

\(\Rightarrow \frac{e}{m}=\frac{v}{rB}\)

• Therefore option 3 is correct.

Important Points

• The relation between kinetic energy (KE) and the radius followed by the charged particle is given by:

\(\Rightarrow r\; = \;\frac{{\sqrt {2m(K E)} }}{{qB}}\)

CONCEPT:

• Faraday's Law of electromagnetic induction states that when a conductor moves inside a Magnetic Field, an electric current induces in the conductor.
• Fleming's right-hand rule:
  • According to Fleming's right-hand rule, the thumb, forefinger, and middle finger of the right hand are stretched to be perpendicular to each other and if the thumb represents the direction of the movement of the conductor, the fore-finger represents the direction of the magnetic field, then the middle finger represents the direction of the induced current.
  • It is used used to determine the direction of current in a generator's windings.

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EXPLANATION:

• From the figure, it is clear that if the forefinger indicates the direction of the magnetic field and the thumb shows the direction of motion of the conductor, then the stretched middle finger will predict the direction of the induced current.

Additional Information

• Fleming's Left-Hand Rule states that if we keep our Fore Finger in the direction of the Magnetic Field, Middle Finger in the direction of the Current then the thumb directs the direction of the Force experienced by the conductor.

Concept:

• The property of matter which is responsible for electrostatic force is called electric charge.
• Proton is a positively charged particle.
• The region around any magnet in which the magnetic force can be experienced by any other magnet or by any magnetic material is called as the magnetic field of the magnet.

Magnetic force on any moving charged particle (F) \(= {\rm{}}q\vec V \times \vec B\)

Where q is a charge, \(\vec V\) its velocity vector, and \(\vec B\) is magnetic field vector

\(\vec V \times \vec B\) is the cross product of velocity vector and magnetic field vector

Explanation:

Magnetic force on any moving charged particle (F) \(= {\rm{}}q\vec V \times \vec B\)

• Since magnetic force is in the direction of the cross product of velocity and magnetic field which is perpendicular to the velocity of the charged particle.

• As the magnetic force on any moving charged particle is perpendicular to the velocity of that particle. So the magnetic force never changes the magnitude of velocity. It just changes the direction of the velocity.
• As the magnitude of the velocity is equal to speed. So the speed of the moving charge particle remains constant in a magnetic field. So option 1 is not correct.
• Mass of the object is also the property that never changes. So option 3 is correct.
• The charge also doesn’t change. So option 2 is wrong.
• As the direction of velocity changes so velocity changes. Hence option 4 is correct.

CONCEPT:

Right-hand thumb rule:

• Right-hand thumb rule states that if a current-carrying conductor is imagined to be held in the right hand such that the thumb points in the direction of the current, then the curled fingers indicate the direction of the magnetic field.
• If the current flows in the upward direction, then the direction of the magnetic field is clockwise, then the direction facing will be the North Pole.
• If the current flows in the downward direction the direction of the magnetic field is anticlockwise, then the direction facing will be the South Pole.

EXPLANATION:

• In the given case thumb points in the east-west direction then according to the right-hand rule, the direction of the magnetic field will be above the conductor i.e., towards the north. 
• Hence option 1 is the answer.

CONCEPT:

• Fleming Left-hand rule gives the force experienced by a charged particle moving in a magnetic field or a current-carrying wire placed in a magnetic field.
• It states that "stretch the thumb, the forefinger, and the central finger of the left hand so that they are mutually perpendicular to each other.
• If the forefinger points in the direction of the magnetic field, the central finger points in the direction of motion of charge, then the thumb points in the direction of force experienced by positively charged particles."

• The magnitude of the force is given by 

⇒ F = I(L × B)

Where I = Amount of current flowing through the conductor, B =  Magnetic field, and L = Length of conductor

EXPLANATION:

• Given that electrons in the beam of a television tube move horizontally from South to North that means current (I) is flowing from North to South.
• The vertical component of the earth's magnetic field points down, which means it is acting in the eastern direction.
• Hence the electrons are deflected toward the West. Therefore option 1 is correct.

CONCEPT:

• Magnetic field: The space or region around the current carrying wire/moving electric charge or around the magnetic material in which force of magnetism can be experienced by other magnetic material is called as magnetic field/magnetic induction by that material/current. It is denoted by B.
• When a current carrying wire is placed in a magnetic field then it experiences a magnetic force on it.

The magnetic force on a current carrying wire is given by:

F = I L B Sin θ

Where I is current in wire, L is length of the wire, B is magnetic field and θ is angle between current and magnetic field.

EXPLANATION:

• Since magnetic force F = I L B Sin θ is force on the current wire in magnetic field. Here the force is directly proportional to the current flowing in the circuit.
• If the current flowing in the circuit is doubled then the force on the wire will also double. So option 1 is correct.