JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
A wire of length \(314\,cm\) carrying current of \(14\,A\) is bent to form a circle. The magnetic moment of the coil is \(........A- m ^{2}\). [Given \(\left.\pi=3.14\right]\)
- A \(10\)
- B \(11\)
- C \(54\)
- D \(0\)
Answer & Solution
Correct Answer
(B) \(11\)
Step-by-step Solution
Detailed explanation
\(\frac{314}{100}=2 \pi R \quad R =0.5 m\) \(\text { Magnetic Moment }= IA\) \(=14 \times \pi R ^{2}\) \(= 14 \times(3.14) \times \frac{1}{4}\)3 \(=10.99 \approx 11.00\)
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