TS EAMCET · Physics · Magnetic Effects of Current
A current carrying loop \(\mathrm{ABCD}\) has two circular arcs \(\mathrm{AD}\) and \(\mathrm{BC}\) with radius \(1 \mathrm{~cm}\) and \(2 \mathrm{~cm}\) respectively as shown in the figure. The two arcs \(\mathrm{AD}\) and \(\mathrm{BC}\) subtend a common angle \(30^{\circ}\) at the centre \(\mathrm{O}\). If the current flowing in the loop is \(\frac{1.2}{\pi} \mathrm{A}\), then the magnitude of net magnetic field at \(O\) is (Given \(\mu_0=4 \pi \times 10^{-7}\) )
- A \(0.5 \mu \mathrm{T}\)
- B \(3 \mu \mathrm{T}\)
- C \(1 \mu \mathrm{T}\)
- D \(1.5 \mu \mathrm{T}\)
Answer & Solution
Correct Answer
(C) \(1 \mu \mathrm{T}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} &\mathrm{B}_0=\mathrm{B}_{\mathrm{AB}}+\mathrm{B}_{\mathrm{BC}}+\mathrm{B}_{\mathrm{CD}}+\mathrm{B}_{\mathrm{DA}} \\ & =\mathrm{O}+\frac{\mu_0 \mathrm{i}}{24 \times 0.02} \otimes+\mathrm{O}+\frac{\mu_0 \mathrm{i}}{24 \times 0.01} \odot \\ & =\frac{\mu_0…
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