The magnetic field due to a current carrying circular loop of radius 3 cm at a point on the axis at a distance of 4 cm from the centre is 54\mu T. What will be its value at the centre of the loop ?

  • Option 1)

    250\mu T

  • Option 2)

    150\mu T

  • Option 3)

    125\mu T

  • Option 4)

    75\mu T

 

Answers (1)
V Vakul

As we discussed in

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

 

 

Magnetic Field due to Circular Current at the centre -

If a coil of radius r is carrying current i then magnetic feild at a distnace X from its centre is:

B_{centre}= \frac{\mu_{0} }{4\pi } \frac{2\pi Ni}{r} = \frac{\mu_{0} Ni}{2r}

- wherein

 

 

 

 

:\; \; Field\; along \; axis\; of \; coil\; \; B= \frac{\mu _{0}iR^{2}}{2\left ( R^{2} +x^{2}\right )^{3/2}}

At \; the\; centre\; of \; coil,\; \; {B}'= \frac{\mu _{0}i}{2R}

\therefore \; \; \; \; \frac{{B}'}{B}= \frac{\mu _{0}i}{2R}\times \frac{2\left ( R^{2}+x^{2} \right )^{3/2}}{\mu _{0}iR^{2}}=\frac{\left ( R^{2}+x^{2} \right )^{3/2}}{R^{3}}

\therefore \; \; \; \; {B}'=\frac{B\times \left ( R^{2}+x^{2} \right )^{3/2}}{R^{3}}

\therefore \; \; \;\; \; \; \; \; \; \; = \frac{54\times \left [ \left ( 3 \right ) ^{2}+ \left ( 4 \right ) ^{2}\right ]^{3/2}}{ \left ( 3 \right ) ^{3}}= \frac{54\times 125}{27}

or \; \; \;\; \; \; {B}'= 250\mu T

 


Option 1)

250\mu T

Correct option

Option 2)

150\mu T

Incorrect option

Option 3)

125\mu T

Incorrect option

Option 4)

75\mu T

Incorrect option

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