A circular loop and an infinitely long straig | vedclass

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

Question

(A) The magnetic field at the centre of a circular loop of radius $r$ carrying current $i$ is $B_{loop} = \frac{\mu_0 i}{2r}$.The magnetic field at a

Vedclass · Accepted Answer

$\frac{15}{29}$

Vedclass · Answer

(A) The magnetic field at the centre of a circular loop of radius $r$ carrying current $i$ is $B_{loop} = \frac{\mu_0 i}{2r}$.The magnetic field at a distance $r$ from an infinitely long straight wire carrying current $i$ is $B_{wire} = \frac{\mu_0 i}{2\pi r}$.When the current in the loop is clockwise,the magnetic field due to the loop is directed into the plane of the paper,and the magnetic field due to the straight wire is directed out of the plane of the paper. Thus,$B_1 = \frac{\mu_0 i}{2r} - \frac{\mu_0 i}{2\pi r} = \frac{\mu_0 i}{2r} (1 - \frac{1}{\pi})$.When the current in the loop is anti-clockwise,the magnetic field due to the loop is directed out of the plane of the paper,and the magnetic field due to the straight wire is also directed out of the plane of the paper. Thus,$B_2 = \frac{\mu_0 i}{2r} + \frac{\mu_0 i}{2\pi r} = \frac{\mu_0 i}{2r} (1 + \frac{1}{\pi})$.Taking the ratio,$\frac{B_1}{B_2} = \frac{1 - \frac{1}{\pi}}{1 + \frac{1}{\pi}} = \frac{\pi - 1}{\pi + 1}$.Substituting $\pi \approx \frac{22}{7}$,we get $\frac{B_1}{B_2} = \frac{\frac{22}{7} - 1}{\frac{22}{7} + 1} = \frac{15}{29}$.

$A$ circular loop and an infinitely long straight conductor carry equal currents,as shown in the figure. The net magnetic field at the centre of the loop is $B_1$,when the current in the loop is clockwise and $B_2$ when the current in the loop is anti-clockwise. Then $\frac{B_1}{B_2}$ is

Similar Questions

$A$ current $I$ enters a circular coil of radius $R$,branches into two parts and then recombines as shown in the circuit diagram. The resultant magnetic field at the centre of the coil is

The magnetic field at the origin due to the current $I$ flowing in the wire as shown in the figure is:

$A$ current $I$ flows in an anticlockwise direction in a circular arc of a wire having $\left(\frac{3}{4}\right)^{\text{th}}$ of the circumference of a circle of radius $R$. The magnetic field $B$ at the centre of the circle is $(\mu_0 = \text{permeability of free space})$

The magnetic field induction produced at the centre of orbit due to an electron revolving in $n^{\text{th}}$ orbit of a hydrogen atom is proportional to

What is the magnetic field at a distance $R$ from a coil of radius $r$ carrying current $I$ along its axis?

Vedclass Test Series

Exam Paper Generator

Online Exam Module