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  • Can someone explain Consider two thin identical conducting wires covered with very thin insulating material. One of the wires is bent into a loop and produces magnetic field B1' at its centre when a current I passes through it. The second wire is

Consider two thin identical conducting wires covered with very thin insulating  material. One of the wires is bent into a loop and produces magnetic field B1'  at its centre when a current I passes through it.  The second wire is bent into a coil with  three identical loops adjacent to each other and produces magnetic field B2 at the centre of the loops when current I/3 passes through it. The ratio B1 : B2 is :  

 

  • Option 1)

    1:1

  • Option 2)

    1:3

  • Option 3)

    1:9

  • Option 4)

    9:1

 

Answers (1)

As we discussed in concept

Magnetic field due to cCrcular Current Carrying arc -

B=\frac{\mu_{o}}{4\pi}\:\frac{2\pi i}{r}=\frac{\mu_{o}i}{2r}

- wherein

 

 and concept

Magnetic Field at the axis due to circular current carrying wire -

B_{axis}=\frac{\mu_{0}}{4\pi }.\frac{2\pi Nir^{2}}{(x^{2}+r^{2})\frac{3}{2}}

- wherein

N is numbers of turn in coil

 

 For loop  B=\frac{\mu _{0}nI}{2a} where a is the radius of the loop.

B_{1}=\frac{\mu _{0}I}{2a}

For co.l        B=\frac{\mu _{0}I}{4\pi}\frac{2nA}{x^{3}}          at the centre x=radius of loop

B_{2}=\frac{\mu _{0}}{4\pi}\frac{2\times 3\times\left ( \frac{5}{3} \right )\times \pi\left ( \frac{a}{3} \right )^{3}}{a^{3}}= \frac{\mu _{0}.33}{2a}

\frac{B_{1}}{B_{2}}= \frac{\frac{\mu_{0} I}{2a}}{\frac{\mu _{0}.3I}{2a}}=\frac{B_{1}}{B_{2}}= 113

 


Option 1)

1:1

Incorrect Option

Option 2)

1:3

Correct Option

Option 3)

1:9

Incorrect Option

Option 4)

9:1

Incorrect Option

Posted by

Sabhrant Ambastha

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