Q205 / 225 · JEE Main 2004 · Physics
The magnetic field due to a current carrying circular loop of radius $3 \mathrm{~cm}$ at a point on the axis at a distance of $4 \mathrm{~cm}$ from the centre is $54 \mu \mathrm{T}$. What will be its value at the centre of the Loop?
- Correct answerA$250 \mu \mathrm{T}$
- B$150 \mu \mathrm{T}$
- C$125 \mu \mathrm{T}$
- D$75 \mu \mathrm{T}$
Solution
Using formula $B=\frac{\mu_0^{\text {II }}}{2\left(R^2+X^2\right)^{3 / 2}}$, we get
$$
54=\frac{\mu_0 i(3)^2}{2\left[(3)^2+(4)^2\right]^{3 / 2}}
$$
At the centre of the coil, $X=0$ and $B=\frac{\mu_0 i}{2(3)}$
Using equation (i)
$$
B=\frac{54 \times 5^3}{(3)^2 \times 3} \Rightarrow B=250 \mu \mathrm{T} \text {. }
$$
Hence, the correct answer is option: 'A'.