MHT CET · Physics · Magnetic Effects of Current
A wire of certain length carries a steady current. It is first bent to form a circular coil of one turn. The same wire is then bent to form a circular coil of three turns. The ratio of magnetic inductions at the centre of the coil in the two cases is
- A 1:9
- B 1:3
- C 3:1
- D 9:1
Answer & Solution
Correct Answer
(A) 1:9
Step-by-step Solution
Detailed explanation
Magnetic induction at the centre of the coil with \(\mathrm{N}\) turns is given by
\(B_{\mathrm{N}}=\left(\frac{\mu_0 I}{2 r}\right) \mathrm{N}\)
Where \(r\) is the radius of each coil is, \(I\) is the current flowing through.
\(\therefore\) for single loop coil \(2 \pi r=L\)
\(B_1=\frac{\mu_0 I(1)}{2\left(\frac{L}{2 \pi}\right)}=\frac{\mu_0 I \pi}{L}\)
\& For the three-loop coil \((2 \pi r) 3=L\)
\(B_3=\frac{\mu_0 I(3)}{2\left(\frac{L}{6 \pi}\right)}=\frac{9 \mu_0 I \pi}{L}\)
Or \(\frac{B_1}{B_2}=\frac{1}{9}\)
\(B_{\mathrm{N}}=\left(\frac{\mu_0 I}{2 r}\right) \mathrm{N}\)
Where \(r\) is the radius of each coil is, \(I\) is the current flowing through.
\(\therefore\) for single loop coil \(2 \pi r=L\)
\(B_1=\frac{\mu_0 I(1)}{2\left(\frac{L}{2 \pi}\right)}=\frac{\mu_0 I \pi}{L}\)
\& For the three-loop coil \((2 \pi r) 3=L\)
\(B_3=\frac{\mu_0 I(3)}{2\left(\frac{L}{6 \pi}\right)}=\frac{9 \mu_0 I \pi}{L}\)
Or \(\frac{B_1}{B_2}=\frac{1}{9}\)
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