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Question

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A. $M{{T}^{-1}}{{C}^{-1}}$

B. $M{{T}^{-2}}{{C}^{-1}}$

C. $M{{T}^{-1}}{{C}^{-1}}$

D. $M{{T}^{-2}}{{C}^{-2}}$

Answer
Verified

Hint: Convert the derived physical quantities into fundamental physical quantities. Magnetic field depends on the force on the charge particle, charge and the velocity of the charge particle. So, we can find the dimensional formula of magnetic field from the dimension of these three quantities.

Formula used:

$F=q\vec{E}+q\vec{v}\times \vec{B}$

Complete Step-by-Step solution:

Lorentz force is the combination of electric and magnetic force on a point charge due to an electromagnetic field. Lorentz force is given by,

$F=q\vec{E}+q\vec{v}\times \vec{B}$

Where,

F is the electromagnetic force, q is the charge, E is the electric field, v is the velocity of the moving charge and B is the magnetic field.

Magnetic force is given by,

$F=q\vec{v}\times \vec{B}$

Magnetic field is,

$B=\dfrac{F}{qv}$

For an equation to be dimensionally correct, dimensions of every term on both sides of the equal sign and related by addition and subtraction should be the same.

Now,

Dimensions of L.H.S. = Dimensions of L.H.S

Dimension of B = dimension of $\dfrac{F}{qv}$

Now, force is a derived quantity. We have to express in terms of the fundamental quantities.

Now, dimension of force is given as,

$\text{Force}=\text{mass}\times \text{acceleration}$

$\text{Acceleration}=\text{displacement}\times \text{tim}{{\text{e}}^{(-2)}}$

Now. dimension of acceleration= $\left[ {{M}^{0}}{{L}^{1}}{{T}^{-2}} \right]$

Dimension of force=$[{{M}^{1}}]\times [{{M}^{0}}{{L}^{1}}{{T}^{-2}}]=[{{M}^{1}}{{L}^{1}}{{T}^{-2}}]$

Dimension of charge, q = $\left[ {{M}^{0}}{{L}^{0}}{{T}^{0}}{{C}^{1}} \right]$

Dimension of velocity, v = $\left[ {{M}^{0}}{{L}^{1}}{{T}^{-1}} \right]$

Now, the dimension of magnetic field B is given by,

$\begin{align}

& \left[ B \right]=\left[ {{M}^{1}}{{L}^{1}}{{T}^{-2}} \right]\times {{\left[ {{M}^{0}}{{L}^{0}}{{T}^{0}}{{C}^{1}} \right]}^{-1}}\times {{\left[ {{M}^{0}}{{L}^{1}}{{T}^{-1}} \right]}^{-1}} \\

& \left[ B \right]=\left[ {{M}^{1}}{{L}^{0}}{{T}^{-1}}{{C}^{-1}} \right] \\

& \left[ B \right]=\left[ {{M}^{1}}{{T}^{-1}}{{C}^{-1}} \right] \\

\end{align}$

The correct option is (C)

Note: Don’t try to remember the dimensional formula. You may get confused. Always express the derived quantities in terms of the fundamental quantities and you will get the dimension of quantities.

Formula used:

$F=q\vec{E}+q\vec{v}\times \vec{B}$

Complete Step-by-Step solution:

Lorentz force is the combination of electric and magnetic force on a point charge due to an electromagnetic field. Lorentz force is given by,

$F=q\vec{E}+q\vec{v}\times \vec{B}$

Where,

F is the electromagnetic force, q is the charge, E is the electric field, v is the velocity of the moving charge and B is the magnetic field.

Magnetic force is given by,

$F=q\vec{v}\times \vec{B}$

Magnetic field is,

$B=\dfrac{F}{qv}$

For an equation to be dimensionally correct, dimensions of every term on both sides of the equal sign and related by addition and subtraction should be the same.

Now,

Dimensions of L.H.S. = Dimensions of L.H.S

Dimension of B = dimension of $\dfrac{F}{qv}$

Now, force is a derived quantity. We have to express in terms of the fundamental quantities.

Now, dimension of force is given as,

$\text{Force}=\text{mass}\times \text{acceleration}$

$\text{Acceleration}=\text{displacement}\times \text{tim}{{\text{e}}^{(-2)}}$

Now. dimension of acceleration= $\left[ {{M}^{0}}{{L}^{1}}{{T}^{-2}} \right]$

Dimension of force=$[{{M}^{1}}]\times [{{M}^{0}}{{L}^{1}}{{T}^{-2}}]=[{{M}^{1}}{{L}^{1}}{{T}^{-2}}]$

Dimension of charge, q = $\left[ {{M}^{0}}{{L}^{0}}{{T}^{0}}{{C}^{1}} \right]$

Dimension of velocity, v = $\left[ {{M}^{0}}{{L}^{1}}{{T}^{-1}} \right]$

Now, the dimension of magnetic field B is given by,

$\begin{align}

& \left[ B \right]=\left[ {{M}^{1}}{{L}^{1}}{{T}^{-2}} \right]\times {{\left[ {{M}^{0}}{{L}^{0}}{{T}^{0}}{{C}^{1}} \right]}^{-1}}\times {{\left[ {{M}^{0}}{{L}^{1}}{{T}^{-1}} \right]}^{-1}} \\

& \left[ B \right]=\left[ {{M}^{1}}{{L}^{0}}{{T}^{-1}}{{C}^{-1}} \right] \\

& \left[ B \right]=\left[ {{M}^{1}}{{T}^{-1}}{{C}^{-1}} \right] \\

\end{align}$

The correct option is (C)

Note: Don’t try to remember the dimensional formula. You may get confused. Always express the derived quantities in terms of the fundamental quantities and you will get the dimension of quantities.

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