A current I enters a circular coil of radius R branches into two parts and then recombines as shown in the circuit diagram.The resultant magnetic field at the centre of the coil is

Question

A current I enters a circular coil of radius R branches into two parts and then recombines as shown in the circuit diagram.The resultant magnetic field at the centre of the coil is (A) zero (B)  (μ 0I/2R)  (C)  (3/4)( (μ 0I/2R) )  (D)  (1/4)( (μ 0I/2R) )

Tardigrade Admin · Accepted Answer

Magnetic field at B B1=(μ 0/2π )(IΔ I/r) Magnetic field at A B2=(μ 0/2π )(IΔ I/r) The resultant magnetic field at the centre |BA|-|BB|=0 So magnetic field becomes zero.

Q. A current $I$ enters a circular coil of radius $R$ branches into two parts and then recombines as shown in the circuit diagram.The resultant magnetic field at the centre of the coil is

$zero$

$\frac{μ _{0} I}{2 R}$

$\frac{3}{4} (\frac{μ _{0} I}{2 R})$

$\frac{1}{4} (\frac{μ _{0} I}{2 R})$

$\frac{1}{2} (\frac{μ _{0} I}{2 R})$

Magnetic field at B
$B_{1} = \frac{μ _{0}}{2 π} \frac{I Δ I}{r}$
Magnetic field at A
$B_{2} = \frac{μ _{0}}{2 π} \frac{I Δ I}{r}$
The resultant magnetic field at the centre
$∣ B_{A} ∣ - ∣ B_{B} ∣ = 0$
So magnetic field becomes zero.

Q. A current I enters a circular coil of radius R branches into two parts and then recombines as shown in the circuit diagram.The resultant magnetic field at the centre of the coil is

zero

2Rμ0​I​

43​(2Rμ0​I​)

41​(2Rμ0​I​)

21​(2Rμ0​I​)

Magnetic field at B B1​=2πμ0​​rIΔI​ Magnetic field at A B2​=2πμ0​​rIΔI​ The resultant magnetic field at the centre ∣BA​∣−∣BB​∣=0 So magnetic field becomes zero.