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  • The magnetic field due to current carrying circular loop of radius 3cm at a point on the axis at a distance of 4cm from the centre is <img alt="54 \mu T" src="https://entrancecorner.oncodecogs

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

Option: 1

250 \mu \mathrm{T}


Option: 2

350 \mu \mathrm{T}


Option: 3

380 \mu \mathrm{T}


Option: 4

200 \mu \mathrm{T}


Answers (1)

best_answer

Magnetic field on the axis of circular conductor,

B=\frac{\mu_0 I}{2} \frac{r^2}{\left(r^2+x^2\right)^{3 / 2}} and magnetic field at the centre (x=0) is,

B^{\prime}=\frac{\mu_0 I}{2 r} 

\therefore \frac{B^{\prime}}{B}=\frac{\left(r^2+x^2\right)^{3 / 2}}{r^3}

\begin{aligned} B^{\prime} & =\frac{\left(r^2+x^2\right)^{3 / 2}}{r^3} B \\ & =\frac{\left(3^2+4^2\right)^{3 / 2}}{3^3} \times 54 \\ & =250 \mu \mathrm{T} \end{aligned}

 

Posted by

Kuldeep Maurya

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