Question

If the resistance of the upper half of a rigid loop is twice that of the lower half, the magnitude of magnetic induction at the centre is equal to

• A. zero
• B. $\frac{{\mu }_{0}I}{4a}$
• C. $\frac{{\mu }_{0}I}{12a}$
• D. $\frac{{\mu }_{0}I}{3a}$

Answer 1

Answer: (C)

$\because I_1{R}_1=I_2{R}_2$
$\frac{I_1}{I_2}=\frac{R_2}{R_1}=\frac{R_2}{2R_2}=\frac{1}{2}$
$\therefore I_2=2I_1$
and $I=I_1+I_2=I_1+2I_1=3I_1$
$\therefore I_1=\frac{I}{3}$ and $I_2=\frac{2}{3}I$
Magnetic field at the centre of coil is
${\vec{B}}_O=B_1\otimes +B_2\odot$
$\therefore B_O=-\frac{{\mu }_0{I}_1\pi }{4\pi a}+\frac{{\mu }_0{I}_2\pi }{4\pi a}$
$B_O=\frac{{\mu }_0\pi }{4\pi a}\left(I_2-I_1\right)$
$=\frac{{\mu }_0}{4a}\left(\frac{2}{3}I-\frac{I}{3}\right)$
$=\frac{{\mu }_0}{4a}\left(\frac{I}{3}\right)=\frac{{\mu }_{0}I}{12a}$