A 150 m long metal wire connects points A and B. The electric potential at point B is 50V less than that at point A. If the conductivity of the metal is `60 xx 10^(6)mho//m` then magnitude of the current density in the wire is equal to:

Options

A. `11 xx 10^(-4)A//m^(2)`
B. `5.5 xx 10^(-3)A//m^(2)`
C. `4 xx 10^(7)A//m^(2)`
D. `20 xx 10^(6)A//m^(2)`

Answer

To solve the problem, we need to find the magnitude of the current density in the wire connecting points A and B. Let's break down the solution step by step.

### Step 1: Understand the given information
We have:
- Length of the wire, \( L = 150 \, \text{m} \)
- Potential difference, \( V = 50 \, \text{V} \) (the potential at point B is 50 V less than at point A)
- Conductivity of the metal, \( \sigma = 60 \times 10^6 \, \text{mho/m} \)

### Step 2: Calculate resistivity
The resistivity \( \rho \) is the reciprocal of conductivity:
\[ \rho = \frac{1}{\sigma} = \frac{1}{60 \times 10^6} \, \Omega \cdot \text{m} \]

### Step 3: Calculate resistance of the wire
The resistance \( R \) of the wire can be calculated using the formula:
\[ R = \rho \frac{L}{A} \]
Substituting the values we have:
\[ R = \left(\frac{1}{60 \times 10^6}\right) \frac{150}{A} \]

### Step 4: Simplify the resistance expression
To simplify:
\[ R = \frac{150}{60 \times 10^6 \cdot A} = \frac{2.5}{10^6 \cdot A} \, \Omega \]

### Step 5: Calculate the current using Ohm's Law
Using Ohm's Law, the current \( I \) can be calculated as:
\[ I = \frac{V}{R} \]
Substituting the values:
\[ I = \frac{50}{\frac{2.5}{10^6 \cdot A}} = 50 \cdot \frac{10^6 \cdot A}{2.5} \]
\[ I = 20 \times 10^6 \, A \]

### Step 6: Calculate current density
Current density \( J \) is defined as:
\[ J = \frac{I}{A} \]
Substituting the expression for current:
\[ J = \frac{20 \times 10^6 \cdot A}{A} = 20 \times 10^6 \, \text{A/m}^2 \]

### Final Answer
The magnitude of the current density in the wire is:
\[ \boxed{20 \times 10^6 \, \text{A/m}^2} \]