Question

The unit of the ratio of magnetic flux to resistance is equal to the unit of which quantity?
A) Charge
B) Potential difference
C) Current
D) Magnetic field

Answer 1

Hint: Magnetic flux refers to the number of field lines passing through a closed surface. It is the dot product of the magnetic field and the area of the given surface. Resistance refers to the opposition offered to the flow of current in a circuit and it is given by the potential difference divided by the current in the circuit. The dimensional analysis will help in determining the unit of the ratio between these two quantities.

Complete step by step solution:
Step 1: Express the dimensional formula of magnetic flux and resistance.

Magnetic flux in a closed surface is given by, ${\varphi }_B=B\cdot A$ where $B$ is the magnetic field and $A$ is the area of the closed surface.

The dimensional formula of the magnetic field is given by, $B\to \left[M{T}^{-2}{I}^{-1}\right]$ where $M$ is the dimension of mass, $T$ is the dimension of time and $I$ is the dimension of current.

The dimensional formula of the area is $A\to \left[L^2\right]$ where $L$ is the dimension of length.

So the dimensional formula of magnetic flux will be

${\varphi }_B\to \left[M{T}^{-2}{I}^{-1}\right]\left[L^2\right]=\left[M{T}^{-2}{I}^{-1}{L}^2\right]$ -------- (1)

The resistance in a circuit is given by, $R={\displaystyle \frac{V}{I}}$ where $V$ is the potential difference and $I$ is the current in the circuit.

Since the dimensional formula for voltage is $V\to \left[M{L}^2{T}^{-3}{I}^{-1}\right]$ and that of current is $I\to \left[I\right]$ , the dimensional formula of resistance will be $R\to {\displaystyle \frac{\left[M{L}^2{T}^{-3}{I}^{-1}\right]}{\left[I\right]}}=\left[M{L}^2{T}^{-3}{I}^{-2}\right]$ --------- (2)


Step 2: Express the given ratio in terms of the dimensional formulas of the two quantities.
We need to determine the unit of the ratio of magnetic flux to resistance i.e., ${\displaystyle \frac{{\varphi }_B}{R}}$ .
To express this ratio in terms of the dimensional formulas of the magnetic flux and resistance we divide equation (1) by (2).

Then we have ${\displaystyle \frac{{\varphi }_B}{R}}\to {\displaystyle \frac{\left[M{T}^{-2}{I}^{-1}{L}^2\right]}{\left[M{L}^2{T}^{-3}{I}^{-2}\right]}}$

Simplifying the powers, we obtain ${\displaystyle \frac{{\varphi }_B}{R}}\to \left[I^1{T}^1\right]$ .

Since the charge is given by, $q=\int Idt$ where $I$ is current and $t$ is the time, its dimensional formula will be $q\to \left[I^1{T}^1\right]$ .

So the ratio of the magnetic flux to resistance has the same dimensional formula as that of the charge and so we can assume that they both have the same units as well.

Hence the correct option is A.

Note: Alternate method
The magnetic flux can also be defined in terms of inductance as ${\varphi }_B=LI$ where $L$ is the self-inductance and $I$ is the current.
Then the ratio of magnetic flux to resistance will be ${\displaystyle \frac{{\varphi }_B}{R}}={\displaystyle \frac{L}{R}}I$ ------ (A)
In equation (A), the term ${\displaystyle \frac{L}{R}}$ refers to the time constant and will thus have the unit of second, the current has the unit of Ampere.
So the unit of the above ratio will be Ampere-second.
Now the unit of charge is Coulomb but this is just Ampere-second as the charge is given by, $q=\int Idt$ where $I$ is current and $t$ is the time.
So the ratio of the magnetic flux to the resistance has the same unit as that of charge.