First slide

Magnetic field due to current element

Question

A current loop, having two circular arcs joined by two radial lines as shown in the figure. It carries a current of 10 A. The magnetic field at point O will be close to

Moderate

A

$1.0 \times 10^{- 5} T$
B

$1.0 \times 10^{- 7} T$
C

$1.5 \times 10^{- 7} T$
D

$1.5 \times 10^{- 5} T$

Solution

From the given figure as shown below
The magnetic field at point O due to wires PQ and RS will be zero.
Magnetic field due to arc QR at point O will be
$B_{1} = \frac{θ}{2 π} (\frac{μ_{0} i}{2 a})$
Here, $θ = 45^{\circ} = \frac{π}{4} rad, i = 10 A$
and $a = 3 cm = 3 \times 10^{- 2} m$
$\begin{matrix} \Rightarrow B_{1} = \frac{π}{2 π \times 4} (\frac{μ_{0} \times 10}{2 \times 3 \times 10^{- 2}}) \\ = \frac{μ_{0} \times 5}{2 \times 12 \times 10^{- 2}} = \frac{5 \times μ_{0} \times 10^{2}}{24} \end{matrix}$
Direction of field B₁ will be coming out of the plane of figure.
Similarly, field at point O due to arc SP will be
$\begin{matrix} B_{2} = \frac{π}{4} (\frac{1}{2 π}) [\frac{μ_{0} \times 10}{2 \times (2 + 3) \times 10^{- 2}}] \\ = \frac{μ_{0} \times 10^{2}}{8} \end{matrix}$
Direction of B₂ is going into the plane of the figure.
$∴$ The resultant field at O is
$\begin{array}{l} B = B_{1} - B_{2} = \frac{1}{2} (\frac{5 \times μ_{0}}{12 \times 10^{- 2}} - \frac{μ_{0}}{4 \times 10^{- 2}}) \\ = \frac{4 π \times 10^{- 7}}{12 \times 10^{- 2}} ≅ 1 \times 10^{- 5} T \end{array}$

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