The dimensions of `h//e` (`h=` Plank's constant and `e=` electronic charge ) are same as that of

Options

A. magnetic field
B. electric flux
C. electric field
D. magnetic field

Answer

To solve the question regarding the dimensions of \( \frac{h}{e} \) (where \( h \) is Planck's constant and \( e \) is the electronic charge), we will follow these steps:

### Step 1: Determine the Dimension of Planck's Constant \( h \)

Planck's constant \( h \) can be expressed in terms of energy and frequency. The formula is:

\[ h = E \cdot t \]

where \( E \) is energy and \( t \) is time. The dimension of energy \( E \) is given by:

\[ [E] = [M][L^2][T^{-2}] \]

Thus, the dimension of \( h \) is:

\[ [h] = [E][t] = [M][L^2][T^{-2}][T] = [M][L^2][T^{-1}] \]

### Step 2: Determine the Dimension of Electric Charge \( e \)

The electric charge \( e \) can be defined in terms of current \( I \) and time \( t \):

\[ e = I \cdot t \]

The dimension of current \( I \) is:

\[ [I] = [A] \]

Thus, the dimension of electric charge \( e \) is:

\[ [e] = [I][t] = [A][T] \]

### Step 3: Calculate the Dimension of \( \frac{h}{e} \)

Now that we have the dimensions of \( h \) and \( e \), we can calculate the dimension of \( \frac{h}{e} \):

\[ \frac{h}{e} = \frac{[M][L^2][T^{-1}]}{[A][T]} = [M][L^2][T^{-1}][A^{-1}] \]

This simplifies to:

\[ \frac{h}{e} = [M][L^2][T^{-2}][A^{-1}] \]

### Step 4: Compare with Given Quantities

Now we need to compare this dimension with the dimensions of the given quantities:

1. **Magnetic Field**:
\[ [B] = [M][L^0][T^{-2}][A^{-1}] \]

2. **Electric Flux**:
\[ [\Phi_E] = [M][L^3][T^{-3}][A^{-1}] \]

3. **Electric Field**:
\[ [E] = [M][L][T^{-3}][A^{-1}] \]

4. **Magnetic Flux**:
\[ [\Phi_B] = [M][L^2][T^{-2}][A^{-1}] \]

### Conclusion

The dimension \( [M][L^2][T^{-2}][A^{-1}] \) of \( \frac{h}{e} \) matches with the dimension of magnetic flux \( [\Phi_B] \).

Thus, the final answer is:

\[ \text{The dimensions of } \frac{h}{e} \text{ are the same as that of magnetic flux.} \]

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