In a particular system of units, a physical quantity can be expressed in terms of the electric charge � �, electron mass 𝑚�, Planck’s constant ℎ, and Coulomb’s constant 𝑘 = � , where 𝜖� is the permittivity of vacuum. In terms of these physical constants, the dimension of the magnetic field is [𝐵] = [𝑒]�[𝑚�]�[ℎ]�[𝑘]�. The value of 𝛼 + 𝛽 + 𝛾 +𝛿 is ____ .

Alright, let's tackle this problem step by step. We need to determine the exponents $\alpha$, $\beta$, $\gamma$, and $\delta$ such that the dimensions of the magnetic field $B$ can be expressed as:

$$[B]=[e{]}^{\alpha }[m_e{]}^{\beta }[h{]}^{\gamma }[k{]}^{\delta }$$

And then find the sum $\alpha +\beta +\gamma +\delta$.

Understanding the Dimensions

First, let's recall the dimensions of each of the given physical constants in terms of mass (M), length (L), time (T), and electric charge (Q):

1. Electric charge (e):

   • $[e]=Q$

2. Electron mass (m_e):

   • $[m_e]=M$

3. Planck’s constant (h):

   • Planck’s constant has units of action, which is energy × time.

   • Energy has dimensions $M{L}^2{T}^{-2}$, so:

     • $[h]=M{L}^2{T}^{-1}$

4. Coulomb’s constant (k):

   • Coulomb's constant is given by $k=\frac{1}{4\pi {ϵ}_0}$.

   • The force between two charges is $F=k\frac{e^2}{r^2}$, so:

     • $[F]=ML{T}^{-2}=[k]\frac{Q^2}{L^2}$

     • Therefore, $[k]=ML{T}^{-2}\cdot L^2\mathrm{/}{Q}^2=M{L}^3{T}^{-2}{Q}^{-2}$

5. Magnetic field (B):

   • The force on a moving charge in a magnetic field is $F=qvB$, so:

     • $[F]=ML{T}^{-2}=Q\cdot L{T}^{-1}\cdot [B]$

     • Therefore, $[B]=ML{T}^{-2}\mathrm{/}(Q\cdot L{T}^{-1})=M{T}^{-1}{Q}^{-1}$

So, the dimension of the magnetic field is:

$$[B]=M{T}^{-1}{Q}^{-1}$$

Setting Up the Dimensional Equation

Now, express $[B]$ in terms of the other constants:

$$[B]=[e{]}^{\alpha }[m_e{]}^{\beta }[h{]}^{\gamma }[k{]}^{\delta }$$

Substituting the dimensions:

$$M{T}^{-1}{Q}^{-1}=Q^{\alpha }\cdot M^{\beta }\cdot (M{L}^2{T}^{-1}{)}^{\gamma }\cdot (M{L}^3{T}^{-2}{Q}^{-2}{)}^{\delta }$$

Let's expand the right-hand side:

$$=Q^{\alpha }\cdot M^{\beta }\cdot M^{\gamma }{L}^{2\gamma }{T}^{-\gamma }\cdot M^{\delta }{L}^{3\delta }{T}^{-2\delta }{Q}^{-2\delta }$$

Combine like terms:

• For $M$: $\beta +\gamma +\delta$

• For $L$: $2\gamma +3\delta$

• For $T$: $-\gamma -2\delta$

• For $Q$: $\alpha -2\delta$

So, we have:

$$M{T}^{-1}{Q}^{-1}=M^{\beta +\gamma +\delta }{L}^{2\gamma +3\delta }{T}^{-\gamma -2\delta }{Q}^{\alpha -2\delta }$$

Equating the Dimensions

Now, equate the exponents for each fundamental dimension on both sides:

1. Mass (M):

   • $1=\beta +\gamma +\delta$

2. Length (L):

   • $0=2\gamma +3\delta$ (since there's no length on the left)

3. Time (T):

   • $-1=-\gamma -2\delta$

4. Charge (Q):

   • $-1=\alpha -2\delta$

Now, we have a system of four equations:

1. $\beta +\gamma +\delta =1$

2. $2\gamma +3\delta =0$

3. $\gamma +2\delta =1$

4. $\alpha -2\delta =-1$

Solving the System of Equations

Let's solve these equations step by step.

From equation 2:

$$2\gamma +3\delta =0\Rightarrow 2\gamma =-3\delta \Rightarrow \gamma =-\frac{3}{2}\delta$$

From equation 3:

$$\gamma +2\delta =1$$

Substitute $\gamma$ from above:

$$\begin{gathered} -\frac{3}{2}\delta +2\delta =1 \\ (-\frac{3}{2}+2)\delta =1 \\ \left(\frac{1}{2}\right)\delta =1 \\ \delta =2 \end{gathered}$$

Now, find $\gamma$:

$$\gamma =-\frac{3}{2}\times 2=-3$$

Now, from equation 1:

$$\begin{gathered} \beta +\gamma +\delta =1 \\ \beta -3+2=1 \\ \beta -1=1 \\ \beta =2 \end{gathered}$$

Finally, from equation 4:

$$\begin{gathered} \alpha -2\delta =-1 \\ \alpha -4=-1 \\ \alpha =3 \end{gathered}$$

Verifying the Solution

Let's check if these values satisfy all equations:

1. $\beta +\gamma +\delta =2+(-3)+2=1$ ✅

2. $2\gamma +3\delta =2(-3)+3(2)=-6+6=0$ ✅

3. $\gamma +2\delta =-3+4=1$ ✅

4. $\alpha -2\delta =3-4=-1$ ✅

All equations are satisfied.

Calculating the Sum

Now, compute $\alpha +\beta +\gamma +\delta$:

$$\alpha +\beta +\gamma +\delta =3+2+(-3)+2=4$$

Final Answer

The value of $\alpha +\beta +\gamma +\delta$ is 4.