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Standard XII

Physics

Ampere's Circuital Law

Using the for...

Question

Using the formula $\vec{F} = q \vec{v} \times \vec{B} a n d B = \frac{μ_{0} i}{2 π r}$ , show that the SI units of the magnetic field B and the permeability constant µ₀ may be written as N mA⁻¹ and NA⁻² respectively.

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Solution

Using the relation $\vec{F} = q \overset{⇀}{v} \times \vec{B}$ , we get
$B = \frac{F}{q v} = \frac{F}{I t v}$
Units of
Force (F) = N
Current (I) = A
Time (T) = s
Velocity (v) = m/s
$\Rightarrow B = N / A - m$
Now, using the relation $B = \frac{μ_{0} i}{2 π r}$ , we get
$\Rightarrow μ_{0} = B \frac{2 π r}{i} = \frac{N}{A - m} \times \frac{m}{A} \Rightarrow μ_{0} = N / A^{2}$

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Similar questions

Using the formula $\to F = \to q v \times \to B$ and $B = \frac{μ_{0} i}{2 π r}$ , show that the SI units of the magnetic field B and the permeability constant $μ_{0}$ may be written as $N m A^{- 1}$ and $N A^{- 2}$ respectively.

Q. Find the dimensions of

\frac{B^{2}}{μ_{0}}

(B = Magnetic field,

μ_{0}

= magnetic permeability).

Q. The dimensional formula of

\frac{1}{2} μ_{0} H^{2}

(

μ_{0}

- Permeability of free space and H-magnetic field intensity) is:

Q. PARAGRAPH
In electromagnetic theory, the electric and magnetic phenomena are related to each other. Therefore, the dimensions of electric and magnetic quantities must also be related to each other. In the questions below,

[E]

and [𝐵] stand for dimensions of electric and magnetic fields respectively, while

[ε_{0}]

and

[μ_{0}]

stand for dimensions of the permittivity and permeability of free space respectively.

[L]

and

[T]

are dimensions of length and time respectively. All the quantities are given in SI units.

The relation between

[ε_{0}]

and

[μ_{0}]

Find the dimensions of

(a) electri field E,

(b) magnetic field B and

The relevant equations are

$F = q E, F = q v B,$ and $B = \frac{μ_{0} I}{2 π a}$

where F is force, q is charge, v is speed, I is current, and a is distance.

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Using the formula F →=qv →×B → and B=μ0i2πr, show that the SI units of the magnetic field B and the permeability constant µ0 may be written as N mA−1 and NA−2 respectively.

Using the relation F →=qv⇀ ×B→, we get B=Fqv =FItv Units of Force (F) = N Current (I) = A Time (T) = s Velocity (v) = m/s ⇒B=N/A-m Now, using the relation B=μ0i2πr, we get ⇒μ0=B2πri=NA-m×mA⇒μ0=N/A2

Using the formula $\vec{F} = q \vec{v} \times \vec{B} a n d B = \frac{μ_{0} i}{2 π r}$ , show that the SI units of the magnetic field B and the permeability constant µ₀ may be written as N mA⁻¹ and NA⁻² respectively.