An arrangement with a pair of quarter circula | vedclass

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

Question

(B) The magnetic field at the center $C$ is the resultant of the magnetic fields produced by the two quarter-circular arcs of radii $r$ and $R$.The ma

Vedclass · Accepted Answer

$\frac{\mu_{0} I}{8} \left(\frac{1}{r} - \frac{1}{R}\right)$ out of the page

Vedclass · Answer

(B) The magnetic field at the center $C$ is the resultant of the magnetic fields produced by the two quarter-circular arcs of radii $r$ and $R$.The magnetic field due to a full circular loop of radius $a$ carrying current $I$ at its center is $B = \frac{\mu_0 I}{2a}$.For a quarter-circular arc, the magnetic field is $B_{arc} = \frac{1}{4} \left(\frac{\mu_0 I}{2a}\right) = \frac{\mu_0 I}{8a}$.Using the right-hand rule:$1$. For the inner arc of radius $r$, the current flows in a counter-clockwise direction, so the magnetic field at $C$ is directed out of the page $(\odot)$.$2$. For the outer arc of radius $R$, the current flows in a clockwise direction, so the magnetic field at $C$ is directed into the page $(\otimes)$.The net magnetic field at $C$ is $B_{net} = B_r - B_R = \frac{\mu_0 I}{8r} - \frac{\mu_0 I}{8R} = \frac{\mu_0 I}{8} \left(\frac{1}{r} - \frac{1}{R}\right)$.Since $r  \frac{1}{R}$, the net magnetic field is directed out of the page.

$(i)$	$(ii)$	$(iii)$
$(A). \frac{\mu_0 i}{r} \otimes$	$(A). \frac{\mu_0 i}{4}(\frac{1}{r_1} - \frac{1}{r_2}) \otimes$	$(A). \frac{\mu_0 i}{4}(\frac{1}{r_1} - \frac{1}{r_2}) \otimes$
$(B). \frac{\mu_0 i}{2r} \odot$	$(B). \frac{\mu_0 i}{4}(\frac{1}{r_1} + \frac{1}{r_2}) \otimes$	$(B). \frac{\mu_0 i}{4}(\frac{1}{r_1} + \frac{1}{r_2}) \otimes$
$(C). \frac{\mu_0 i}{4r} \otimes$	$(C). \frac{\mu_0 i}{4}(\frac{1}{r_1} - \frac{1}{r_2}) \odot$	$(C). \frac{\mu_0 i}{4}(\frac{1}{r_1} - \frac{1}{r_2}) \odot$
$(D). \frac{\mu_0 i}{4r} \odot$	$(D). 0$	$(D). 0$

An arrangement with a pair of quarter circular coils of radii $r$ and $R$ with a common centre $C$ and carrying a current $I$ is shown in the figure. The permeability of free space is $\mu_0$. The magnetic field at $C$ is

Similar Questions

Equal current $i$ is flowing in three infinitely long wires along positive $x, y$ and $z$ directions. The magnetic field at a point $(0, 0, -a)$ would be:

When the current flowing in a circular coil is doubled and the number of turns of the coil in it is halved,the magnetic field at its centre will become

$A$ coil of $50$ turns and $4$ cm radius carries $2$ $A$ current. The magnetic field at its centre is ....... $mT$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module