Find the current density as a funciton of distance `r` from the axis of a radially symmetrical parallel stream of electrons if the magnetic induction inside the streams varies as `B = br^(alpha)`, where `b and `alpha` are positive constants.

Explanation

To find the current density \( j(r) \) as a function of distance \( r \) from the axis of a radially symmetrical parallel stream of electrons, given that the magnetic induction \( B \) varies as \( B = b r^{\alpha} \), we can follow these steps:

### Step 1: Use Ampere's Law
According to Ampere's law, the magnetic field \( B \) around a current-carrying conductor is related to the current density \( j \) by the equation:
\[ B_{\phi} \cdot 2\pi r = \mu_0 \int_0^r j(r') \cdot 2\pi r' \, dr' \]
where \( B_{\phi} \) is the magnetic field at a distance \( r \) from the axis, \( \mu_0 \) is the permeability of free space, and \( j(r') \) is the current density at a distance \( r' \).

### Step 2: Substitute the Given Magnetic Field
From the problem, we know that:
\[ B_{\phi} = b r^{\alpha} \]
Substituting this into Ampere's law gives:
\[ (b r^{\alpha}) \cdot 2\pi r = \mu_0 \int_0^r j(r') \cdot 2\pi r' \, dr' \]
The \( 2\pi \) cancels out on both sides:
\[ b r^{\alpha + 1} = \mu_0 \int_0^r j(r') r' \, dr' \]

### Step 3: Differentiate Both Sides
To find \( j(r) \), we differentiate both sides with respect to \( r \):
\[ \frac{d}{dr}(b r^{\alpha + 1}) = \frac{d}{dr}\left(\mu_0 \int_0^r j(r') r' \, dr'\right) \]
Using the Fundamental Theorem of Calculus on the right side:
\[ (\alpha + 1) b r^{\alpha} = \mu_0 j(r) r \]

### Step 4: Solve for Current Density \( j(r) \)
Rearranging the equation to isolate \( j(r) \):
\[ j(r) = \frac{(\alpha + 1) b r^{\alpha}}{\mu_0 r} \]
This simplifies to:
\[ j(r) = \frac{(\alpha + 1) b}{\mu_0} r^{\alpha - 1} \]

### Final Result
Thus, the current density as a function of distance \( r \) from the axis is:
\[ j(r) = \frac{(\alpha + 1) b}{\mu_0} r^{\alpha - 1} \]

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