The current through the circular coil is halved and the radius of the coil is doubled. If $B_1$ and $B_2$ are respectively the initial and final magnetic field strength, then:

A galvanometer of resistance $50 \Omega$ is having 30 divisions and a current sensitivity $10 \mathrm{~mA} /$ div. What should be the shunt resistance so that it can be converted into an ammeter of range 10 A ?

Two long parallel conductors $X$ and $Y$ are placed vertically at a distance $d$ apart. Conductor X carries a current I upwards and conductor Y carries a current 21 downwards. A third long conductor Z is placed parallel to both X and Y and between X and Y . If Z carries a current I upwards and is at a distance $x$ from conductor X , then:

A. All the conductors are in equilibrium and the net force on Z is zero

B. The net force on Z is $\frac{\mu_0 I^2}{2 x \pi}\left[\frac{d+x}{d-x}\right]$ towards X

C. The net force on Z is $\frac{\mu_0 I^2}{2 x \pi}\left[\frac{d+x}{d-x}\right]$ towards Y

D. The net force on Z is $\frac{\mu_0 I^2}{2 x \pi}\left[\frac{d-x}{d+x}\right]$ towards Y

A $50 \Omega$ galvanometer is shunted by a resistance of $S \Omega$. If $8 \%$ of total current passes through the galvanometer, the value of $S$ is: