Question

Figure shows two long conducting wires placed at right angles. They carry current as indicated. Assume that I A current in a long wire produces magnetic field B at a distance d from the wire. The strength and direction of field at a point x marked in the figure will be

• A. 3B into the plane of the paper
• B. B into the plane of the paper
• C. B out of the plane of the paper
• D. 3B out of the plane of the paper

Answer 1

The correct option is A 3B into the plane of the paper

Given: current in horizontal wire,$I_1=1A$ in $+xdi{r}^n$

current in vertical wire,$I_2=2A$ in $-ydi{r}^n$

magnetic field at $x$ due to horizontal wire is,$B_1=B$ into the plane of paper

To find: magnetic field at $x=?$

Solution: As we know that,

magnetic field by a long conducting wire at distance a is,

$B=\frac{{\mu }_{0}I}{2\pi a}$

So, Due to horizontal wire :

$B_1=\frac{{\mu }_0{I}_1}{2\pi d}$

$==>B_1=\frac{{\mu }_0}{2\pi d}=B,$ into the plane of paper

and Due to vertical wire :

$B_2=\frac{{\mu }_0{I}_2}{2\pi d}$

$==>B_2=\frac{2{\mu }_0}{2\pi d}=2B,$ into the plane of paper

Net magnetic field at $x:$

$B^{\prime }=B_1+B_2$

$B^{\prime }=B+2B$

$B^{\prime }=3B,$ into the plane of paper

hence,

The correct opt: A