AP EAMCET · PHYSICS · Magnetic Effects of Current
The magnetic field due to a current carrying loop of radius \(3 \mathrm{~cm}\) at a point on its axis at a distance of \(4 \mathrm{~cm}\) from its centre of \(54 \mu \mathrm{T}\). Then, the value of the magnetic field at the centre of the loop is
- A \(250 \mu \top\)
- B \(150 \mu \top\)
- C \(75 \mu \top\)
- D \(125 \mu \top\)
Answer & Solution
Correct Answer
(A) \(250 \mu \top\)
Step-by-step Solution
Detailed explanation
\(B_{\text {axis }}=54 \times 10^{-6}=\frac{\mu N I r^2}{2\left(r^2+x^2\right)^{3 / 2}}\) Here, \(\quad r=3 \mathrm{~cm}=3 \times 10^{-2} \mathrm{~m}\) and \(x=4 \mathrm{~cm}=4 \times 10^{-2} \mathrm{~m}\). From above relation, we get…
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