What is the dimensional formula of magnetic flux/electric flux?

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Jasmine Grover

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Ques: What is the dimensional formula of magnetic flux/electric flux?

Ans: In physics, the dimensional formula of magnetic flux is given by:

Magnetic Flux = [M] [L]2 [T]-2 [I]-1

Where:

also, the dimensional formula of electric flux is given by:

[Electric Flux] = [M] [L]3 [T]-3 [I]-1

Therefore, the dimensional formula of magnetic flux to electric flux can be obtained by dividing the dimensional formula of magnetic flux by the dimensional formula of electric flux. 

Thus, the dimensional formula of Magnetic flux/Electric flux = [M] [L]2 [T]-2 [I]-1/[M] [L]3 [T]-3 [I]-1 = [L]-1 [T]

Important Facts about Electric Flux & Magnetic Flux

  • Magnetic flux is a physical quantity which can be defined as a measure of the strength of a magnetic field passing through a given area. It is used to describe the behaviour of magnetic fields in different materials.
  • Electric flux is a measure of the strength of an electric field passing through a given area. It is calculated by taking the dot product of the electric field vector and the area vector.
  • The ratio of magnetic flux to electric flux is measured in inverse length and time.
  • The ratio of electric flux to magnetic flux is given by Electric flux/magnetic flux = [M] [L]3 [T]-3 [I]-1/[M] [L]2 [T]-2 [I]-1
  • Dimensional of electric flux to magnetic flux is LT−1

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CBSE CLASS XII Related Questions

  • 1.
    Nuclides with the same number of neutrons are called:

      • Isobars
      • Isotones
      • Isotopes
      • Isomers

    • 2.
      A circular coil of 100 turns and radius \( \left(\frac{10}{\sqrt{\pi}}\right) \, \text{cm}\) carrying current of \( 5.0 \, \text{A} \) is suspended vertically in a uniform horizontal magnetic field of \( 2.0 \, \text{T} \). The field makes an angle \( 30^\circ \) with the normal to the coil. Calculate:
      the magnetic dipole moment of the coil, and
      the magnitude of the counter torque that must be applied to prevent the coil from turning.


        • 3.
          A square loop of side 0.50 m is placed in a uniform magnetic field of 0.4 T perpendicular to the plane of the loop. The loop is rotated through an angle of 60° in 0.2 s. The value of emf induced in the loop will be:

            • 5 V
            • 3.5 V
            • 2.5 V
            • Zero V

          • 4.
            Four long straight thin wires are held vertically at the corners A, B, C and D of a square of side \( a \), kept on a table and carry equal current \( I \). The wire at A carries current in upward direction whereas the current in the remaining wires flows in downward direction. The net magnetic field at the centre of the square will have the magnitude:

              • \( \dfrac{\mu_0 I}{\pi a} \) and directed along OC
              • \( \dfrac{\mu_0 I}{\pi a \sqrt{2}} \) and directed along OD
              • \( \dfrac{\mu_0 I \sqrt{2}}{\pi a} \) and directed along OB
              • \( \dfrac{2\mu_0 I}{\pi a} \) and directed along OA

            • 5.
              Two small identical metallic balls having charges \( q \) and \( -2q \) are kept far at a separation \( r \). They are brought in contact and then separated at distance \( \frac{r}{2} \). Compared to the initial force \( F \), they will now:

                • attract with a force \( \frac{F}{2} \)
                • repel with a force \( \frac{F}{2} \)
                • repel with a force \( F \)
                • attract with a force \( F \)

              • 6.
                Light of which of the following colours will have the maximum energy in a photon associated with it?

                  • Red light
                  • Yellow light
                  • Green light
                  • Blue light
                CBSE CLASS XII Previous Year Papers

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