$[M^1L^0T^{-2}A^{-1}]$
To find the dimensional formula for the magnetic flux density \( B \), we need to start from the Lorentz force equation:
\(\vec{F} = q(\vec{v} \times \vec{B})\),
where \( \vec{F} \) is the force, \( q \) is the electric charge, \( \vec{v} \) is velocity, and \( \vec{B} \) is the magnetic flux density.
We need to establish the relationship between these variables in terms of their dimensions:
Using the Lorentz force equation, equate the dimensions of both sides:
[M^1L^1T^{-2}] = [A^1T^1] \cdot [L^1T^{-1}] \cdot [B]
Rearrange to find the dimension of \( B \):
[B] = \frac{[M^1L^1T^{-2}]}{[A^1T^1] \cdot [L^1T^{-1}]}
Simplify the expression,
[B] = \frac{[M^1L^1T^{-2}]}{[A^1L^1T^0]}
Cancel out and simplify:
[B] = [M^1L^0T^{-2}A^{-1}]
Thus, the dimensional formula for magnetic flux density \( B \) is correctly given by [M^1L^0T^{-2}A^{-1}].
Therefore, the correct answer is the option:
[M^1L^0T^{-2}A^{-1}].
Given that the energy stored in an inductor is expressed as $U_L = \frac{1}{2}LI^2$ and the power dissipated in a resistor is $P = I^2R$, where $L$ is inductance, $R$ is resistance, and $I$ is current, determine the dimension of the ratio $\frac{L}{R}$.
The dimensions of energy are:
If force $[F]$, acceleration $[A]$ and time $[T]$ are chosen as the fundamental physical quantities. Find the dimensions of pressure.