A horizontal semi-circular wire of radius $r$ is connected to a battery through two similar springs $X$ and $Y$ to an electric cell, which sends current $I$ through it. A vertically downward uniform magnetic field $B$ is applied on the wire, as shown in the figure. What is the force acting on each spring?
To determine the force acting on each spring, we need to find the total magnetic force acting on the semi-circular wire and then consider how that force is distributed between the two springs.
Total Magnetic Force on the Semi-circular Wire:
The magnetic force on a current-carrying wire in a magnetic field is given by \( \vec{F} = I \int d\vec{l} \times \vec{B} \). For a straight wire of length \(L\) perpendicular to the magnetic field, the magnitude of the force is simply \(F = BIL\). However, we have a curved wire.
Instead of integrating, we can calculate the magnetic force on the curved semi-circular wire by considering the effective length. The effective length is the straight-line distance between the endpoints of the curved wire. In this case, the endpoints are separated by a distance equal to the diameter of the semi-circle, which is \(2r\).
Therefore, the total magnetic force \(F\) acting on the semi-circular wire is: \[F = B I (2r) = 2BIr\] The direction of this force is outward from the center of the semicircle, due to the right-hand rule.
Force on Each Spring:
Since there are two identical springs \(X\) and \(Y\) supporting the wire, and the magnetic force acts along the diameter of the semicircle, we can assume that the force is equally distributed between the two springs.
Therefore, the force acting on each spring is half of the total magnetic force: \[F_{\text{spring}} = \frac{F}{2} = \frac{2BIr}{2} = BIr\]
Thus, the force acting on each spring is \(BIr\).
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