First slide

Force on a current carrying wire placed in a magnetic field

Question

A long horizontal wire P carries a current of 50 A. It is rigidly fixed. Another fine wire Q is placed directly above and parallel to P. The weight of wire Q is 0.075 N/m and carries a current of 25 A. Find the position of wire Q from P so that the wire Q remains suspended due to magnetic repulsion.

Moderate

A

$10^{- 2} m$
B

$\frac{10^{- 2}}{2} m$
C

$\frac{10^{- 2}}{3} m$
D

$\frac{10^{- 2}}{4} m$

Solution

As force per unit length between two parallel current carrying wires separated by a distance dis given by $\frac{dF}{dl} = \frac{_{μ_{0}}}{4 π} \frac{2 i_{1} i_{2}}{d}$
It is repulsive if the current in the wires are in opposite direction.
In order that wire Q remains suspended, the magnetic force must be equal to its weight.
$\begin{array}{l} So, F_{m} = Mg \\ \frac{F}{L} = \frac{Mg}{L} \Rightarrow \frac{μ_{0} i_{1} i_{2}}{2 πd} = \frac{Mg}{L} \\ d = \frac{μ_{0} i_{1} i_{2}}{2 π \frac{Mg}{L}} = \frac{4 π \times 10^{- 7} \times 50 \times 25}{2 π \times 0 .075} m \\ d = \frac{10^{- 2}}{3} m \end{array}$

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