A very long conducting wire is bent in a semi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The magnetic field at point $P$ is the sum of the fields due to the two semi-infinite straight wires and the semicircular arc.$1$. The magnetic fi

Vedclass · Accepted Answer

$\frac{\mu_0 i }{4 R }\left[1-\frac{2}{\pi}\right]$ pointed away from the page

Vedclass · Answer

(D) The magnetic field at point $P$ is the sum of the fields due to the two semi-infinite straight wires and the semicircular arc.$1$. The magnetic field due to each semi-infinite wire at distance $R$ is $B_{straight} = \frac{\mu_0 i}{4 \pi R}$. Using the right-hand rule,both wires produce a magnetic field pointing into the page.$2$. The magnetic field due to the semicircular arc of radius $R$ is $B_{arc} = \frac{\mu_0 i}{4 R}$. Using the right-hand rule,this field points out of the page.$3$. The net magnetic field is $B_{net} = B_{arc} - 2 	imes B_{straight} = \frac{\mu_0 i}{4 R} - 2 \left( \frac{\mu_0 i}{4 \pi R} ight) = \frac{\mu_0 i}{4 R} \left( 1 - \frac{2}{\pi} ight)$.Since $1 > \frac{2}{\pi}$,the net field points out of the page (away from the page).

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

Similar Questions

The magnetic induction at the centre $O$ in the figure shown is

Two long wires carrying currents of $8 \,A$ and $6 \,A$ are placed along the $x$-axis and $y$-axis respectively. Find the magnitude of the magnetic field at the point $P(2, 4)$. (Take $\mu_{0} = 4\pi \times 10^{-7} \,T \cdot m/A$)

Define the law for finding the direction of a magnetic field due to a circular current loop.

Vedclass Test Series

Exam Paper Generator

Online Exam Module