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

$A$ steady current $I$ flows through a wire with one end at $O$ and the other end extending up to infinity as shown in the figure. The magnetic field at a point $P$, located at a distance $d$ from $O$, is

Two long straight parallel conductors $10 \text{ cm}$ apart,carry equal currents of magnitude $3 \text{ A}$ in the same direction. Then,the magnetic induction at a point midway between them is

Magnetic field intensity at the centre of a coil of $50$ turns,radius $0.5\, m$ and carrying a current of $2\, A$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module