One of the two identical conducting wires of length L is bent in the form of a circular loop and the other one into a circular coil of N identical turns. If the same current is passed in both, the ratio of the magnetic field at the central of the loop (BL) to that at the centre of the coil (BC) , i.e., R (BL/BC) will be

Question

One of the two identical conducting wires of length L is bent in the form of a circular loop and the other one into a circular coil of N identical turns. If the same current is passed in both, the ratio of the magnetic field at the central of the loop (BL) to that at the centre of the coil (BC) , i.e., R (BL/BC) will be (A) (1/N) (B) N2 (C) (1/N2) (D) N

Tardigrade Admin · Accepted Answer

L = 2 π R L = N ×2π r R =Nr BL = (μ0 i/2R) BC = (μ0 Ni/2r) BC = (μ0 N2 i/2R) (BL/BC) = (1/N2)

$\frac{1}{N}$

$N^{2}$

$\frac{1}{N ^{2}}$

$N$

$L = 2 π R$ $L = N \times 2 π r$
$R = N r$
$B_{L} = \frac{μ _{0} i}{2 R}$
$B_{C} = \frac{μ _{0} N i}{2 r} B_{C} = \frac{μ _{0} N ^{2} i}{2 R}$
$\frac{B _{L}}{B _{C}} = \frac{1}{N ^{2}}$

N1​

N2

N21​

N

L=2πR L=N×2πr R=Nr BL​=2Rμ0​i​ BC​=2rμ0​Ni​BC​=2Rμ0​N2i​ BC​BL​​=N21​

$\frac{1}{N}$

$N^{2}$

$\frac{1}{N ^{2}}$

$N$

$L = 2 π R$ $L = N \times 2 π r$
$R = N r$
$B_{L} = \frac{μ _{0} i}{2 R}$
$B_{C} = \frac{μ _{0} N i}{2 r} B_{C} = \frac{μ _{0} N ^{2} i}{2 R}$
$\frac{B _{L}}{B _{C}} = \frac{1}{N ^{2}}$