A conducting circular loop of copper is place | vedclass

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

Question

(A) The resistance of a wire is given by $R = \frac{\rho L}{A}$.For part $abc$,the length is $L_{abc} = \frac{3}{4}(2 \pi R) = \frac{3 \pi R}{2}$ and

Vedclass · Accepted Answer

$\frac{\mu_0 I}{8 R} \otimes$

Vedclass · Answer

(A) The resistance of a wire is given by $R = \frac{ho L}{A}$.For part $abc$,the length is $L_{abc} = \frac{3}{4}(2 \pi R) = \frac{3 \pi R}{2}$ and area is $A$. So,$R_{abc} = \frac{ho (3 \pi R / 2)}{A} = \frac{3 \pi ho R}{2A}$.For part $adc$,the length is $L_{adc} = \frac{1}{4}(2 \pi R) = \frac{\pi R}{2}$ and area is $A/3$. So,$R_{adc} = \frac{ho (\pi R / 2)}{A/3} = \frac{3 \pi ho R}{2A}$.Since $R_{abc} = R_{adc}$,the current $I$ divides equally into $I/2$ in each part.The magnetic field due to a circular arc at its center is $B = \frac{\mu_0 I 	heta}{4 \pi R}$.For part $abc$,$	heta = 3 \pi / 2$,so $B_{abc} = \frac{\mu_0 (I/2) (3 \pi / 2)}{4 \pi R} = \frac{3 \mu_0 I}{16 R}$ (directed into the page,$\otimes$).For part $adc$,$	heta = \pi / 2$,so $B_{adc} = \frac{\mu_0 (I/2) (\pi / 2)}{4 \pi R} = \frac{\mu_0 I}{16 R}$ (directed out of the page,$\odot$).The net magnetic field is $B_{net} = B_{abc} - B_{adc} = \frac{3 \mu_0 I}{16 R} - \frac{\mu_0 I}{16 R} = \frac{2 \mu_0 I}{16 R} = \frac{\mu_0 I}{8 R}$ (directed into the page,$\otimes$).

$A$ conducting circular loop of copper is placed as shown in the figure. The cross-sectional area of part $abc$ is $A$ and that of part $adc$ is $A/3$. The magnetic field at point $O$ is:

Similar Questions

An electric current is flowing through a circular coil of radius $R$. The ratio of the magnetic field at the centre of the coil and that at a distance $2\sqrt{2}R$ from the centre of the coil on its axis is:

What will be the resultant magnetic field at the origin due to four infinite length wires? If each wire produces a magnetic field '$B$' at the origin.

$A$ current $i$ flows in a circular arc of wire of radius $R$,which subtends an angle of $3\pi / 2$ radians at its centre. The magnetic induction at the centre is

$A$ horizontal overhead powerline is at a height of $4\ m$ from the ground and carries a current of $100\ A$ from east to west. The magnetic field directly below it on the ground is: (Given: $\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A$)

There are two infinitely long straight current-carrying conductors held at right angles to each other such that their common ends meet at the origin, as shown in the figure. The ratio of current in both conductors is $1:1$. The magnetic field at point $P(x, y)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module