Question

Two circular coils 1 &2 are made from the same wire but the radius of the1st coil is twice that of the 2nd coil.what is the ratio of potential difference in volt should be applied across them so that the magnetic field at their centres is the same

Answer 1

"Let r1 and r2 are the radius of coil 1 & 2. If B1 and B2 are magnetic induction at their centre, then

B1=μ0I12r1&B2=μ0I22r2

Since B1 = B2; and r1 = 2r2 therefore I1 = 2I2 Again if R1 and R2 are resistance of the coil 1 and 2 then R1 = 2R2 (as R ∝ length = 2πr) and if V1 and V2 are the potential difference across them respectively, then

V1V2=I1R1I2R2 = (2I2)(2R2)I2R2=4

"

Answer 2

Answer:4 times

Explanation:

Magnetic field due to circular coil at centre

B_{Centre}=\frac{\mu_{0}}{4\pi }\frac{2\pi Ni}{r}=\frac{\mu_{0} Ni}{2r}

given B1=B2

i_1/r_1 =i_2/r_2 \\given \ r_1=2r_2 \\ \ so\ i_1/i_2 =2

if R1 and R2 are resistance of the coil 1 and 2 then R1 = 2R2 (as R ∝ length )

so V1/V2 =4