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 –
\[\begin{align}
& \text{A) }\dfrac{{{\mu }_{0}}I}{8R} \\
& \text{B) }\dfrac{{{\mu }_{0}}I}{10R} \\
& \text{C) }\dfrac{{{\mu }_{0}}I}{4R} \\
& \text{D) }\dfrac{{{\mu }_{0}}I}{12R} \\
\end{align}\]
Answer
642k+ views
Hint: 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 –
\[\begin{align}
& \phi ={{90}^{0}} \\
& \Rightarrow \sin {{90}^{0}}=1 \\
\end{align}\]
Also, we have the length of the arc from the from the radius and the angle subtended as –
\[\begin{align}
& l=R\theta \\
& \text{here,} \\
& \theta \text{=6}{{\text{0}}^{0}}=\dfrac{\pi }{3} \\
& \Rightarrow l=\dfrac{\pi R}{3} \\
\end{align}\]
Now, let us substitute this in the formula for the magnetic field at the point O as –
\[\begin{align}
& B=\dfrac{{{\mu }_{0}}Il\sin \phi }{4\pi {{r}^{2}}} \\
& \text{Substituting all the data,} \\
& B=\dfrac{{{\mu }_{0}}I\dfrac{\pi R}{3}(1)}{4\pi {{R}^{2}}} \\
& \therefore B=\dfrac{{{\mu }_{0}}I}{12R} \\
\end{align}\]
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.
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 –
\[\begin{align}
& \phi ={{90}^{0}} \\
& \Rightarrow \sin {{90}^{0}}=1 \\
\end{align}\]
Also, we have the length of the arc from the from the radius and the angle subtended as –
\[\begin{align}
& l=R\theta \\
& \text{here,} \\
& \theta \text{=6}{{\text{0}}^{0}}=\dfrac{\pi }{3} \\
& \Rightarrow l=\dfrac{\pi R}{3} \\
\end{align}\]
Now, let us substitute this in the formula for the magnetic field at the point O as –
\[\begin{align}
& B=\dfrac{{{\mu }_{0}}Il\sin \phi }{4\pi {{r}^{2}}} \\
& \text{Substituting all the data,} \\
& B=\dfrac{{{\mu }_{0}}I\dfrac{\pi R}{3}(1)}{4\pi {{R}^{2}}} \\
& \therefore B=\dfrac{{{\mu }_{0}}I}{12R} \\
\end{align}\]
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.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

