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Find the dimensions of \(\frac{B^{2}}{\mu_{o}}\) (A) \(M L^{2} T^{-2}\) (B) \(M L^{-1} T^{-1}\) (C) \(M L^{-2} T^{-2}\) (D) \(M L^{-1} T^{-2}\)

Short Answer

Expert verified
The dimensions of \(\frac{B^{2}}{\mu_{o}}\) are \(M L^{2} T^{-2}\). So, the correct option is (A).

Step by step solution

01

Find the dimensions of \(B^{2}\)

The dimensions of magnetic field B (Tesla) are \(M T^{-2} I^{-1}\), so the dimensions of the square of magnetic field (\(B^{2}\)) are \(M^{2} T^{-4} I^{-2}\).
02

Find the dimensions of \(\mu_{o}\)

The permeability of free space \(\mu_{o}\) (Henry per meter) has dimensions \(L M^{-1} T^2 I^2\). The dimension is found from the relation \(F = \mu_{o} I^{2} / 2\pi d\), where F is the force (Newton), I is the current (Ampere) and d is the distance (meter).
03

Determine the dimensions of \(\frac{B^{2}}{\mu_{o}}\)

Now, we divide the dimensions of \(B^{2}\) with the dimensions of \(\mu_{o}\). So, the dimensions of \(\frac{B^{2}}{\mu_{o}}\) would be \(\frac{M^{2} T^{-4} I^{-2}}{L M^{-1} T^2 I^2}\), which simplifies to \(M L^{2} T^{-2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnetic Field
The magnetic field is a fundamental concept in physics. It represents the region around a magnetic material or a moving electric charge where the force of magnetism acts. The strength and direction of a magnetic field can be represented by magnetic field lines. The SI unit for measuring the magnetic field is the Tesla, which is symbolized as \( B \).

In dimensional analysis, the dimensions of a magnetic field are \( M T^{-2} I^{-1} \), where:
  • \( M \) stands for mass (in kilograms)
  • \( T \) stands for time (in seconds)
  • \( I \) stands for electrical current (in amperes)
When analyzing problems involving magnetic fields, these dimensions allow us to understand how variations might change the field's characteristics. It's like having a recipe that explains how different quantities mix.
Permeability of Free Space
The permeability of free space, denoted as \( \mu_{o} \), is a critical constant in electromagnetism. It represents how much resistance the vacuum of space offers to the formation of a magnetic field. The SI unit for permeability of free space is henry per meter \(( \, \text{H/m} \,)\).

Dimensionally, it is represented by \( L M^{-1} T^2 I^2 \). Here:
  • \( L \) is for length (in meters)
  • \( M \) is for mass (in kilograms)
  • \( T \) is for time (in seconds)
  • \( I \) is for electrical current (in amperes)
This dimension tells us the fundamental nature of how a magnetic field can form in free space. It is essential in describing how magnetic and electric fields interact. It underpins the equations that explain electromagnetism fundamentals, like how currents influence magnetic fields.
Dimensions of Physical Quantities
In physics, understanding the dimensions of physical quantities helps clarify their relationships and ensure the equations we use make sense. Each physical quantity, like length or time, has specific dimensions that relate to its measurement.

By breaking down complex equations into their dimensional components, we can see whether or not the equations balance. For instance, if one side of an equation is in terms of meters (length), the other side must also end up in meters for the equation to be consistent.
  • Example: The equation \( E = mc^2 \) requires energy \( E \) to have dimensions of mass \( m \) and speed of light \( c \) squared. This equates to mass times length squared per time squared.
Through these dimensions, we gain insights not just into the mathematical correctness of physical laws but their practical application. It helps in deriving consistent and meaningful scientific predictions and guides us in finding or confirming the right scientific relationships.

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Most popular questions from this chapter

A man running on a horizontal road at \(8 \mathrm{~ms}^{-1}\) finds rain falling vertically. If he increases his speed to \(12 \mathrm{~ms}^{-1}\), he finds that drops make \(30^{\circ}\) angle with the vertical. Find velocity of rain with respect to the road. (A) \(4 \sqrt{7} \mathrm{~ms}^{-1}\) (B) \(8 \sqrt{2} \mathrm{~ms}^{-1}\) (C) \(7 \sqrt{3} \mathrm{~ms}^{-1}\) (D) \(8 \mathrm{~ms}^{-1}\)

A man is moving on his bike with \(54 \mathrm{kmh}^{-1}\). He takes a u-turn in \(10 \mathrm{~s}\) and continues to move with the some velocity. Find average acceleration during this time. (A) \(3.0 \mathrm{~ms}^{-2}\) (B) \(1.5 \mathrm{~ms}^{-2}\) (C) 0 (D) \(-1.5 \mathrm{~ms}^{-2}\)

The pitch of a screw gauge is \(1 \mathrm{~mm}\) and there are 100 divisions on its circular scale. When nothing is put in between its jaws, the zero of the circular scale lies 4 divisions below the reference line. When a steel wire is placed between the jaws, two main scale divisions are clearly visible and 67 divisions on the circular scale are observed. The diameter of the wire is (A) \(2.71 \mathrm{~mm}\) (B) \(2.67 \mathrm{~mm}\) (C) \(2.63 \mathrm{~mm}\) (D) \(2.65 \mathrm{~mm}\)

A man goes \(100 \mathrm{~m}\) north then \(100 \mathrm{~m}\) east and then \(20 \mathrm{~m}\) north and then \(100 \sqrt{2} \mathrm{~m}\) south-West. Find the displacement. (A) \(20 \mathrm{~m}\) West (B) \(20 \mathrm{~m}\) East (C) \(20 \mathrm{~m}\) North (D) \(20 \mathrm{~m}\) South

The respective number of significant figure for the numbers \(23.023,0.0003\) and \(2.1 \times 10^{-3}\) are \(\quad\) [2010] (A) \(4,4,2\) (B) \(5,1,2\) (C) \(5,1,5\) (D) \(5,5,2\)

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