Question

The magnitude of magnetic field due to current carrying arc of radius R, having a current I subtending at an angle of ${{60}^{0}}$ at the center O is –

& \text{A) }\dfrac{{{\mu }_{0}}I}{8R} \\\ & \text{B) }\dfrac{{{\mu }_{0}}I}{10R} \\\ & \text{C) }\dfrac{{{\mu }_{0}}I}{4R} \\\ & \text{D) }\dfrac{{{\mu }_{0}}I}{12R} \\\ \end{aligned}$$

Answer 1

We are given an arc of wire carrying the current through it. We can find the magnetic field at any point from this wire given the distance from this arc to the point. Here, we are given the angle subtended and the radius of the arc under consideration.

Complete answer:
We need to understand the relation of the current carrying element and the magnetic field developed due to this at some point away from the element. The Biot-Savart’s law states that the magnetic field developed at a point ‘r’ distance from the current carrying element is proportional to the current flowing through the element, the length of the element, the sine of the angle between the element and the point and inversely proportional to the square of the distance ‘r’.
i.e.,
$B=\dfrac{{{\mu }_{0}}Il\sin \phi }{4\pi {{r}^{2}}}$
Now, let us consider the given situation. Here, the angle between the point O and the current is –

& \phi ={{90}^{0}} \\\ & \Rightarrow \sin {{90}^{0}}=1 \\\ \end{aligned}$$  Also, we have the length of the arc from the from the radius and the angle subtended as – $$\begin{aligned} & l=R\theta \\\ & \text{here,} \\\ & \theta \text{=6}{{\text{0}}^{0}}=\dfrac{\pi }{3} \\\ & \Rightarrow l=\dfrac{\pi R}{3} \\\ \end{aligned}$$ Now, let us substitute this in the formula for the magnetic field at the point O as – $$\begin{aligned} & B=\dfrac{{{\mu }_{0}}Il\sin \phi }{4\pi {{r}^{2}}} \\\ & \text{Substituting all the data,} \\\ & B=\dfrac{{{\mu }_{0}}I\dfrac{\pi R}{3}(1)}{4\pi {{R}^{2}}} \\\ & \therefore B=\dfrac{{{\mu }_{0}}I}{12R} \\\ \end{aligned}$$ From the above calculations and substitutions we understand that the magnetic field due to a current carrying arc subtended by an angle of $${{60}^{0}}$$ at the center O is found to be $$B=\dfrac{{{\mu }_{0}}I}{12R}$$ **The correct answer is option D.** **Note:** The magnetic field induced at a point by a current carrying element is dependent on the length of the current carrying element as always. Here, we had to find the arc length using the angle and radius information for the magnetic field to be found at the center.