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Q.

Figure shows a part of an electric circuit. ABCD is a rectangular loop made of uniform wire. The length AD=BC= 1cm. The sides AB and DC are long as compared to the other two sides. The magnetic force per unit length acting on the wire DC due to the wire AB if the ammeter reads is 10 A is $x \times 10^{- 4} {Nm}^{- 1}$ . then find the value of $x$ ?

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answer is 5.

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Detailed Solution

By symmetry, each of the wires and carries a current of 5A. As the separation between them is 1cm , the magnetic force per unit length of DC is
$\begin{array}{l} \frac{dF}{dl} = \frac{μ_{0} i_{1} i_{2}}{2 πd} = \frac{(2 \times 10^{- 7} {TmA}^{- 1}) (5 A) (5 A)}{1 \times 10^{- 2} m} \\ = 5 \times 10^{- 4} TA = 5 \times 10^{- 4} {Nm}^{- 1} . \end{array}$

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Figure shows a part of an electric circuit. ABCD is a rectangular loop made of uniform wire. The length AD=BC= 1cm. The sides AB and DC are long as compared to the other two sides. The magnetic force per unit length acting on the wire DC due to the wire AB if the ammeter reads is 10 A is x×10−4Nm−1. then find the value of x?