A certain length of insulated wire can be bent to form either a single circular loop (case I) or a double loop of smaller radius (case II). When the same steady current is passed through the wire, the ratio of the magnetic field at the centre in case I to that in case Il is

Question

A certain length of insulated wire can be bent to form either a single circular loop (case I) or a double loop of smaller radius (case II). When the same steady current is passed through the wire, the ratio of the magnetic field at the centre in case I to that in case Il is (A) 1 (B) 2 (C) 1/2 (D) 1/4

Tardigrade Admin · Accepted Answer

Magnetic field at the centre of a current carrying circular wire,

B = \frac{μ _{0} n I}{2 r}

Let length of the wire be L.
Case I

2 π r_{1} = L, r_{1} = \frac{L}{2 π} ∴ B_{1} = \frac{μ _{0} I}{2 ( L /2 π )}

Case II

2 \times (2 π r_{2}) = L, r_{2} = \frac{L}{4 π} ∴ B_{2} = \frac{μ _{0} ( 2 I )}{2 ( L /4 π )}

Hence,

\frac{B _{1}}{B _{2}} = \frac{1}{4}

Q. A certain length of insulated wire can be bent to form either a single circular loop (caseI) or a double loop of smaller radius (caseII). When the same steady current is passed through the wire, the ratio of the magnetic field at the centre in case I to that in case Il is

1

2

1/2

1/4

Magnetic field at the centre of a current carrying circular wire, B=2rμ0​nI​ Let length of the wire be L. Case I 2πr1​=L,r1​=2πL​∴B1​=2(L/2π)μ0​I​ Case II 2×(2πr2​)=L,r2​=4πL​∴B2​=2(L/4π)μ0​(2I)​ Hence, B2​B1​​=41​

Q. A certain length of insulated wire can be bent to form either a single circular loop $(c a se I)$ or a double loop of smaller radius $(c a se II)$ . When the same steady current is passed through the wire, the ratio of the magnetic field at the centre in case I to that in case Il is