A wire loop formed by joining two semicircular wires of radii $R_{1}$ and $R_{2}$ carries a current $I$ as shown in the following diagram. The magnetic induction at the centre $O$ is :

$\frac{μ_{0} I π}{4 π R_{1}}$
$\frac{μ_{0} I π}{4 π R_{2}}$
$\frac{μ_{0} I π}{4 π} (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
$\frac{μ_{0} I π}{4 π} (\frac{1}{R_{1}} + \frac{1}{R_{2}})$

Question

A wire loop formed by joining two semicircular wires of radii $R_{1}$ and $R_{2}$ carries a current $I$ as shown in the following diagram. The magnetic induction at the centre $O$ is :

$\frac{μ_{0} I π}{4 π R_{1}}$
$\frac{μ_{0} I π}{4 π R_{2}}$
$\frac{μ_{0} I π}{4 π} (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
$\frac{μ_{0} I π}{4 π} (\frac{1}{R_{1}} + \frac{1}{R_{2}})$

Toppr Admin · Accepted Answer

Magnetic field-xA0- at O due to segments FA and -xA0-CD is equal to zero-Magnetic field at O due to semi-circle ABC-12-x3BC-0I2R2-x2212-k-Magnetic field at O due to semi-circle DEF-12-x3BC-0I2R1-x2212-k-x2192-Bnet-12-x3BC-0I2R2-x2212-k-12-x3BC-0I2R1-x2212-k-x3BC-0I4-1R1-1R2-x2212-k-x3BC-0I-x3C0-4-x3C0-1R1-1R2-x2212-k-where -xA0-k is-xA0- out of the plane of the loop and perpendicular to the loop

A wire loop formed by joining two semicircular wires of radii R1 and R2 carries a current I as shown in the following diagram. The magnetic induction at the centre O is :μ0Iπ4πR1μ0Iπ4πR2μ0Iπ4π(1R1−1R2)μ0Iπ4π(1R1+1R2)