JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
A hairpin like shape as shown in figure is made by bending a long current carrying wire. What is the magnitude of a magnetic field at point \(P\) which lies on the centre of the semicircle ?
- A \(\frac{\mu_{0} I }{4 \pi r }(2-\pi)\)
- B \(\frac{\mu_{0} I }{4 \pi r }(2+\pi)\)
- C \(\frac{\mu_{0} I }{2 \pi r }(2+\pi)\)
- D \(\frac{\mu_{0} I }{2 \pi r }(2-\pi)\)
Answer & Solution
Correct Answer
(B) \(\frac{\mu_{0} I }{4 \pi r }(2+\pi)\)
Step-by-step Solution
Detailed explanation
\(B =2 \times B _{ \text { straight.wire }}+ B _{\text {loop }}\) \(B =2 \times \frac{\mu_{0} i }{4 \pi r }+\frac{\mu_{0} i }{2 r }\left(\frac{\pi}{2 \pi}\right)\) \(B =\frac{\mu_{0} i }{4 \pi r }(2+\pi)\)
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