in a wire AB of uniform cross section in which a potential gradient of 0.1 V/m exist.If the balancing point is foung to be 180 cm from A then what is the Emf of the cell?
To find the EMF of the cell, we need to use the concept of potential difference and the formula for balancing potential.
The balancing point is the point in the wire where the potential difference due to the cell exactly balances the potential difference due to the gradient across the wire.
Given information:
- Potential gradient across the wire, ∆V/∆x = 0.1 V/m
- Balancing point distance, d = 180 cm = 1.8 m
To find the EMF of the cell, we can use the formula for potential difference in a wire:
∆V = E - (∆V/∆x) * d
Where:
- ∆V is the potential difference in the wire
- E is the EMF of the cell
- ∆x is the change in position along the wire
- d is the distance from the reference point
In this case, since the balancing point is at 180 cm from point A, we can say that ∆x = 1.8 m - 0 m = 1.8 m (assuming A as the reference point).
So, substituting the given values into the formula, we can solve for E:
∆V = E - (0.1 V/m) * 1.8 m
Since the potential difference ∆V across the wire is equal to the potential gradient due to the cell, we can set ∆V equal to 0.1 V:
0.1 V = E - (0.1 V/m) * 1.8 m
Simplifying the equation:
0.1 V = E - 0.18 V
Rearranging the equation to solve for E:
E = 0.1 V + 0.18 V
E = 0.28 V
Therefore, the EMF of the cell is 0.28 V.