Question

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 ?

• A. $125\mu T$
• B. $150\mu T$
• C. $250\mu T$
• D. $75\mu T$

Answer 1

Answer: (C)

$250\mu T$

Field along axis of coil $B=\frac{{\mu }_{0}i{R}^2}{2(R^2+X^2{)}^{3/2}}$

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

$\frac{B^{,}}{B}=\frac{{\mu }_{0}i}{2R}\times \frac{2(R^2+x^2{)}^{3/2}}{{\mu }_{0}i{R}^2}=\frac{(R^2+X^2{)}^{3/2}}{R^3}$

$B^{,}=\frac{B\times (R^2+X^2{)}^{3/2}}{R^3}=\frac{54\times [3^2+4^2{]}^{3/2}}{3^3}=\frac{54\times 125}{27}=250\mu T$