Force Between Two Parallel Currents MCQ Quiz - Objective Question with Answer for Force Between Two Parallel Currents - Download Free PDF

Concept:

The magnetic force between two long, straight, parallel current-carrying conductors is governed by Ampere's Circuital Law and the Biot-Savart Law. The force per unit length is given by:

Here,

• F is the force between the conductors
• L is the length over which the force is considered
• μ₀ is the permeability of free space
• I₁ and I₂ are the currents in the two wires
• d is the distance between the conductors

Explanation:

The total magnetic force on one conductor of length L due to the other is derived by multiplying the force per unit length with the length:

This shows that the total force F is directly proportional to the product of the two currents (I₁I₂) and the length (L) of the conductors, and inversely proportional to the distance (d) between them.

Thus,

The correct option is (3).

Calculation:

B = \(\frac{\mu_{0}(5)}{2 \pi \times .01}-\frac{\mu_{0} 4}{2 \pi \times 0.04}\)

= - \(\frac{100 \mu_{0}}{4 \pi}\)

= - 100 × 10^–7

= - 1 × 10^–5 T  

Explanation:

Problem Statement Analysis:

Two parallel current-carrying conductors attract each other with a force of 3 N when placed 2 cm apart. The currents in both conductors are initially the same, and the length of the conductors is constant. If the currents in both conductors are doubled and the distance between them is halved, what will be the new force?

To solve this problem, we'll use the formula for the force per unit length between two parallel current-carrying conductors:

Formula:

The force per unit length (F/L) between two parallel conductors carrying currents I₁ and I₂, separated by a distance d, is given by:

F/L = (μ₀ / 2π) * (I₁ * I₂ / d)

Where:

• μ₀ = permeability of free space (4π × 10^-7 T·m/A)
• I₁ and I₂ = currents in the conductors
• d = distance between the conductors

Initial Conditions:

• Force (F) = 3 N
• Distance (d) = 2 cm = 0.02 m
• Currents (I₁ = I₂) = I

From the initial conditions, the initial force per unit length (F_initial) can be written as:

F_initial = (μ₀ / 2π) * (I * I / 0.02)

New Conditions:

• New currents (I'₁ = I'₂) = 2I
• New distance (d') = 0.02 m / 2 = 0.01 m

The new force per unit length (F_new) can be written as:

F_new = (μ₀ / 2π) * (2I * 2I / 0.01)

Which simplifies to:

F_new = (μ₀ / 2π) * (4I² / 0.01)

Since we know that F_initial = (μ₀ / 2π) * (I² / 0.02), we can express F_new in terms of F_initial:

F_new = 4 * (0.02 / 0.01) * F_initial

F_new = 4 * 2 * F_initial

F_new = 8 * F_initial

Given that F_initial = 3 N, we have:

F_new = 8 * 3 N = 24 N

Therefore, the new force between the conductors is 24 N.

The correct option is: Option 2: 24 

Concept:

F = (μ₀ / 2π) (I₁ I₂) / d

F' = (μ₀ / 2π) (2I₁ I₂) / d = 2F

• The force per unit length between two parallel current-carrying wires is given by Ampère’s force law:
• If the currents flow in opposite directions, the force is repulsive.
• If the currents flow in the same direction, the force is attractive.
• Doubling the current in one wire changes the force proportionally:
• Since the current is also reversed, the force changes from repulsive to attractive.

Calculation:

Two long parallel wires carrying equal currents in opposite directions repel each other with force F.

One wire’s current is doubled and reversed, and we need to find the new force between them.

⇒ Initially, force between wires:

F = (μ₀ / 2π) (I I) / d

⇒ When one current is doubled (2I) and reversed:

F' = (μ₀ / 2π) (2I I) / d

F' = 2F

The new force between the wires is 2F and attractive.

Calculation:

F = \(\frac{\mu_0}{4 \pi} \frac{2 \mathrm{I}_1 \mathrm{I}_2}{\mathrm{r}} l=10^{-3} \mathrm{~N}\)

When current in both the wires in doubled, then

F' = \(\frac{\mu_0}{4 \pi} \frac{2\left(2 \mathrm{I}_1 × 2 \mathrm{I}_2\right)}{\mathrm{r}} l\) = 4 × 10^-3 N

CONCEPT:

• Magnetic field (B): 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 magnetic field/magnetic induction by that material/current.

The magnetic force per unit length between two parallel wires is given by:
\(\frac{F}{l} = \frac{{{\mu _o}}}{{2\pi }}\frac{{2{I_1}{I_2}}}{d}\)

Where μ0 = permittivity of free space, I1 = current in a first wire, I2 = current in a second wire,d = distance between two wires, and l = length of current-carrying wire.

EXPLANATION:

The magnetic force per unit length between two parallel wires is given by
\(\frac{F}{l} = \frac{{{\mu _o}}}{{4\pi }}\frac{{2{I_1}{I_2}}}{d}\)

Here F ∝ \(\frac{1}{d}\)

• If the distance between the conductors is doubled then force will become half of the previous value.
• Hence option 3 is correct.

• If the conductor carries current in the same direction, then the force between them will be attractive.
• If the conductor carries current in the opposite direction, then the force between them will be repulsive

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.
• The magnetic force per unit length between two parallel wires is given by;

\(\frac{F}{l} = \frac{{{\mu _o}}}{{4\pi }}\frac{{2{I_1}{I_2}}}{d}\)

Where μ₀ = permittivity of free space, I₁ = current in a first wire, I₂ = current in a second wire,d = distance between two wires, and l = length of current-carrying wire.

Explanation:

The magnetic force per unit length between two parallel wires is given by;

\(\frac{F}{l} = \frac{{{\mu _o}}}{{4\pi }}\frac{{2{I_1}{I_2}}}{d}\)

• From the above, it is clear that the force between the two parallel wires depends on the medium in which two-wire is placed, the distance between two-wire and the current in wire 1 and wire 2.
• The force between the two parallel wires is independent of the radius of wire, so there will be no impact on the force between the two parallel wires. Hence option 3 is correct.

CONCEPT:

• Magnetic field (B): 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.
• The force of attraction between two parallel conductors is given by

\(F = \frac{\mu_{0}}{4\pi} \frac{I_{1} I_{2}}{2\pi d}\)

Where I₁ = Current on the first conductor, I₂ = Current on the second conductor, and d = Distance between the conductors 

EXPLANATION :

• The parallel conductors which are carrying current in the opposite direction repel because the direction of the magnetic field turns out to be opposite direction hence they repel each other
• Hence, option 2 is the answer

• If the conductor carries the current in the same direction, then the force between them will be attractive.

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.
• The magnetic force per unit length between two parallel wires is given by;

\(F = \;\frac{{{\mu _{0\;\;\;}}{I_1}\;{I_2}}}{{2\pi \;d}}\)

Where μ₀ = permittivity of free space, I₁ = current in the first wire, I₂ = current in the second wire, and d = distance between two wires.

EXPLANATION:

• Since the force per unit length is given by \(F = \;\frac{{{\mu _{0}}{I_1}\;{I_2}}}{{2\pi \;d}}\)
• As the current in wires increases then the force will also increase. So statement 1 is correct.
• As the distance between two wires increases then the force will decrease. So statement 2 is correct.

• F orce is independent of the radius of the wire. So statement 3 is wrong.

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.
• The magnetic force per unit length between two parallel wires is given by;

\(\frac{F}{l} = \frac{{{\mu _o}}}{{2\pi }}\frac{{{I_1}{I_2}}}{d}\)

Where μ0 = permittivity of free space, I1 = current in a first wire, I2 = current in a second wire,d = distance between two wires, and l = length of current-carrying wire.

EXPLANATION:​ 

• The magnetic force per unit length between two parallel wires is given by;

\(\frac{F}{l} = \frac{{{\mu _o}}}{{2\pi }}\frac{{{I_1}{I_2}}}{d}\)

• If the conductor carries the current in the same direction, then the force between them will be attractive.
• If the conductor carries the current in the opposite direction, then the force between them will be repulsive. Therefore option 2 is correct.

Concept:

• Ampere's Circuital Law: It states the relationship between the current and the magnetic field created by it.
• This law states that the integral of magnetic field density (B) along an imaginary closed path is equal to the product of current enclosed by the path and permeability of the medium.

From the Ampere’s circuital law, the magnitude of the field due to the first conductor can be given by,

The force on a segment of length L of the conductor 2 due to the conductor 1 can be given as,

\(B=\frac{{{\mu }_{0}}{{I}_{1}}}{2\pi d}\)

Because of this magnetic field, the force exerted due to magnetic field of conductor 1 on conductor 2 can be expressed as

\(F={{B}_{1}}{{I}_{2}}L=\frac{{{\mu }_{0}}{{I}_{2}}{{I}_{1}}}{2\pi d}L\)

If the current is flowing in the opposite direction, the force between them is repulsive. But the magnitude in both cases is the same.

Here,

F = force exerted

I₁ = amount of current in wire 1

_I2 = amount of current in wire 2

L = length of wire

d = distance of separation

Explanation:

From the above explanation, we can see that the force exerted by two current-carrying wire can be expressed as

 \(F=\frac{{{\mu }_{0}}{{I}_{2}}{{I}_{1}}}{2\pi d}L\)

And if both conductors carrying the same current I the above equation can be expressed as

 \(F=\frac{{{\mu }_{0}}{{I}^{2}}}{2\pi d}L\)

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.
• The magnetic force per unit length between two parallel wires is given by;

\(\frac{F}{l} = \frac{{{\mu _o}}}{{2\pi }}\frac{{2{I_1}{I_2}}}{d}\)

Where μ0 = permittivity of free space, I1 = current in a first wire, I2 = current in a second wire,d = distance between two wires, and l = length of current-carrying wire.

EXPLANATION:

​Given - I₁ = I₂ = I and distance the between the two-wire (d) = r

• The magnetic force per unit length between two parallel wires is given by;

\(\Rightarrow \frac{F}{l} = \frac{{{\mu _o}}}{{4\pi }}\frac{{2{I_1}{I_2}}}{d}\)

\(\Rightarrow \frac{F}{l} = \frac{{{\mu _o}}}{{4\pi }}\frac{{2{I^2}{}}}{r}=\frac{\mu_o}{2\pi}\frac{I^2}{r}\)

• As the current in the wire is in the opposite direction, then the force between them will be repulsive.

Important Points

• If the conductor carries the current in the same direction, then the force between them will be attractive.
• If the conductor carries the current in the opposite direction, then the force between them will be repulsive

CONCEPT:

• Ampere's Circuital Law: It states the relationship between the current and the magnetic field created by it.
• This law states that the integral of magnetic field density (B) along an imaginary closed path is equal to the product of current enclosed by the path and permeability of the medium.

From the Ampere’s circuital law, the magnitude of the field due to the first conductor can be given by,

B_a = μ₀ I₁ / 2 π d

The force on a segment of length L of the conductor 2 due to the conductor 1 can be given as,

F₂₁ = I₂ L B₁ = μ₀ I₁ I₂ L / 2πd

EXPLANATION:

• From the equation
  \({{\rm{F}}_{21}} = \frac{{{{\rm{\mu }}_{0{\rm{\;}}}}{{\rm{I}}_1}{\rm{\;}}{{\rm{I}}_2}{\rm{\;L}}}}{{2{\rm{\;\pi \;d}}}}\)

So we can see that if the distance is increasing between the conductors then the force will decrease because it is inversely proportional to distance. So option 2 is correct.

CONCEPT:

Magnetic field:

• The space or region around the current-carrying wire/moving electric charge or around a magnet 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.

The magnetic force per unit length between two parallel wires is given by;

\(F = \frac{{{\mu _{0\;}}{I_1}\;{I_2}}}{{2\pi \;d}}\)

Where μ₀ is the permittivity of free space, I₁ is d.c current in first wire and I₂ is d.c current in second wire and d is the distance between two wires.

EXPLANATION:

• When two wires carrying a d.c current are placed parallel to each other then there always will be a magnetic force acting between two wires. So options 3 and 4 are wrong.
• When the current flowing in the wires are in the same direction then the force between two wires is attractive. So option 1 is correct and option 2 is wrong.
• When current is in opposite direction then force is repulsive.

CONCEPT:

• Ampere's Circuital Law: It states the relationship between the current and the magnetic field created by it.
• This law states that the integral of magnetic field density (B) along an imaginary closed path is equal to the product of current enclosed by the path and permeability of the medium.

From  Ampere’s circuital law, the magnitude of the field due to the first conductor can be given by,

Ba = μ0 I1 / 2 π d

The force on a segment of length L of conductor 2 due to conductor 1 can be given as,

F21 = I2 L B1 = μ0 I1 I2 L / 2πd

EXPLANATION:

• From the equation
  \({{\rm{F}}_{21}} = \frac{{{{\rm{\mu }}_{0{\rm{\;}}}}{{\rm{I}}_1}{\rm{\;}}{{\rm{I}}_2}{\rm{\;L}}}}{{2{\rm{\;\pi \;d}}}}\)
• Since \(I_1=I_2=I\)
•  
• F=\(\rm \frac{2\times 10^{-7}I^2}{ x} \frac{N}{m}\)