To solve the problem, we need to find the dimensions of the electric field (E), magnetic field (B), and permeability (μ), and then use these dimensions to determine the value of \(\frac{(b - c)}{a}\).
### Step-by-Step Solution:
1. **Determine the dimension of the electric field (E)**:
- The electric field \(E\) is defined as force per unit charge.
- The unit of force is Newton (N), which can be expressed as \(kg \cdot m/s^2\).
- The unit of charge is Coulomb (C).
- Therefore, the dimension of electric field \(E\) can be expressed as:
\[
E = \frac{N}{C} = \frac{kg \cdot m/s^2}{C} = \frac{M \cdot L}{T^2 \cdot A}
\]
- Thus, the dimension of \(E\) is:
\[
[E] = M^1 L^1 T^{-2} A^{-1}
\]
2. **Determine the dimension of the magnetic field (B)**:
- The magnetic field \(B\) is defined as force per unit charge per unit velocity.
- The unit of magnetic field is Tesla (T), which can also be expressed as \(N/(C \cdot m/s)\).
- Therefore, the dimension of magnetic field \(B\) can be expressed as:
\[
B = \frac{N}{C \cdot (m/s)} = \frac{kg \cdot m/s^2}{C \cdot m/s} = \frac{kg}{C \cdot s}
\]
- Thus, the dimension of \(B\) is:
\[
[B] = M^1 L^0 T^{-2} A^{-1}
\]
3. **Determine the dimension of permeability (μ)**:
- The permeability \(\mu\) is defined in terms of magnetic field and electric field.
- The unit of permeability is Henry (H), which can be expressed as \(N \cdot s^2/C^2\).
- Therefore, the dimension of permeability \(\mu\) can be expressed as:
\[
\mu = \frac{N \cdot s^2}{C^2} = \frac{kg \cdot m/s^2 \cdot s^2}{C^2} = \frac{kg \cdot m}{C^2}
\]
- Thus, the dimension of \(\mu\) is:
\[
[\mu] = M^1 L^1 T^{-2} A^{-2}
\]
4. **Combine the dimensions of \(E\), \(B\), and \(\mu\)**:
- We need to find the dimension of \(\frac{EB}{\mu}\):
\[
\frac{[E][B]}{[\mu]} = \frac{(M^1 L^1 T^{-2} A^{-1})(M^1 L^0 T^{-2} A^{-1})}{(M^1 L^1 T^{-2} A^{-2})}
\]
- This simplifies to:
\[
= \frac{M^{1+1} L^{1+0} T^{-2-2} A^{-1-1}}{M^1 L^1 T^{-2} A^{-2}} = \frac{M^2 L^1 T^{-4} A^{-2}}{M^1 L^1 T^{-2} A^{-2}}
\]
- Canceling out the common dimensions gives:
\[
= M^{2-1} L^{1-1} T^{-4+2} A^{-2+2} = M^1 L^0 T^{-2} A^0
\]
- Thus, the final dimension is:
\[
= M^1 T^{-2}
\]
5. **Compare with the given dimension \(M^a L^b T^c\)**:
- From our result, we have \(a = 1\), \(b = 0\), and \(c = -2\).
6. **Calculate \(\frac{(b - c)}{a}\)**:
- Substitute the values:
\[
\frac{(b - c)}{a} = \frac{(0 - (-2))}{1} = \frac{2}{1} = 2
\]
### Final Answer:
\[
\frac{(b - c)}{a} = 2
\]
