Magnetic Effects Of Current Question 20
Question: A length L of wire carries a steady current I. It is bent first to form a circular plane coil of one turn. The same length is now bent more sharply to give a double loop of smaller radius. The magnetic field at the centre caused by the same current is [NCERT 1980; AIIMS 1980; MP PMT 1995, 99]
Options:
A) A quarter of its first value
B) Unaltered
C) Four times of its first value
D) A half of its first value
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Answer:
Correct Answer: C
Solution:
Magnetic field at the centre of current carrying coil is given by $ B=\frac{{\mu_{0}}}{4\pi }\cdot \frac{2\pi Ni}{r} $
Þ $ B\propto \frac{N}{r} $
Þ $ \frac{B_{1}}{B_{2}}=\frac{N_{1}}{{N_{2}}}\times \frac{r_{2}}{r_{1}} $ . The following figure shows that single turn coil changes to double turn coil. N1 = 1 N2 = 2 r1 = r r2 = r / 2 B1 = B B2 =?
Þ $ \frac{B}{B_{2}}=\frac{1}{2}\times \frac{r/2}{r}=\frac{1}{4} $
Þ $ B_{2}=4B $ Short trick: For such type of problems remember $ B_{2}=n^{2}B_{1} $
BETA
