A very long conducting wire is bent in a semi- circular shape from $A$ to $B$ as shown in figure. The magnetic field at point $P$ for steady current configuration is given by:

Question

A very long conducting wire is bent in a semi- circular shape from A to B as shown in figure. The magnetic field at point P for steady current configuration is given by:

Aicharya · Accepted Answer

Let's consider the following diagram:

The formula to calculate the magnetic field $({\vec{B}}_{1})$ due to wire 1 is given by

${\vec{B}}_{1} = \frac{μ_{0} i}{4 π R} \otimes . . . (1)$

The formula to calculate the magnetic field $({\vec{B}}_{2})$ due to wire 2 is given by

${\vec{B}}_{2} = \frac{μ_{0} i}{4 R} ⊙ . . . (2)$

And, the formula to calculate the magnetic field $({\vec{B}}_{3})$ due to wire 3 is given by

${\vec{B}}_{3} = \frac{μ_{0} i}{4 π R} \otimes . . . (3)$

Hence, the net magnetic field at P can be written as

$\begin{array}{rcl} \vec{B} & = & {\vec{B}}_{1} + {\vec{B}}_{2} + {\vec{B}}_{3} \\ = & \frac{μ_{0} i}{4 π R} \otimes + \frac{μ_{0} i}{4 R} ⊙ + \frac{μ_{0} i}{4 π R} \otimes \\ = & \frac{μ_{0} i}{2 π R} \otimes + \frac{μ_{0} i}{4 R} ⊙ \\ = & - \frac{μ_{0} i}{2 π R} ⊙ + \frac{μ_{0} i}{4 R} ⊙ \\ = & \frac{μ_{0} i}{4 R} [1 - \frac{2}{π}] ⊙ \end{array}$

A very long conducting wire is bent in a semi- circular shape from A to B as shown in figure. The magnetic…

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