Question

The position of point from wire ‘B’. where net magnetic field is zero due to following current distribution.:
(a)

A. $\dfrac{6}{7}\,cm$
B. \[\dfrac{{12}}{7}\,cm\]
C. \[\dfrac{{18}}{7}\,cm\]
D. \[\dfrac{{16}}{7}\,cm\]

(b)

A. $4\,cm$
B. $2\,cm$
C. $8\,cm$
D. $12\,cm$

Answer 1

Hint:A magnetic field is a vector field that describes the magnetic influence on moving electric charges, electric current and magnetic materials. A moving charge in a magnetic field experiences a force perpendicular to its own velocity and to the magnetic field. The SI unit is Tesla ‘T’.

Formula used:
$B = \dfrac{{\mu I}}{{2\pi r}}$
Here, $I$=current and $r$=distance.

Complete step by step answer:
(a)
 

For ${B_{net}} = 0$
And ${B_1} = {B_2}$
As we know that
$\dfrac{{\mu {I_1}}}{{2\pi {r_1}}} = \dfrac{{\mu {I_2}}}{{2\pi {r_2}}}$
Put the values
\[\dfrac{{\mu \times (5i)}}{{2\pi \times x}} = \dfrac{{\mu \times (2i)}}{{2\pi \times (6 - x)}}\]
Simplify
\[(5i)(6 - x) = (2i)(x)\]
\[\Rightarrow 30 - 5x = 2x\]
\[\Rightarrow x = \dfrac{{30}}{7}\,cm\]
From B,
$\Rightarrow (6 - x)$
Put the values
$\Rightarrow (6 - \dfrac{{30}}{7})$
$\Rightarrow \dfrac{{12}}{7}\,cm$

So the correct answer is option B.

(b)

For
${B_{net}} = 0$
And ${B_1} = {B_2}$
As we know that
$\dfrac{{\mu {I_1}}}{{2\pi {r_1}}} = \dfrac{{\mu {I_2}}}{{2\pi {r_2}}}$
Put the values
\[\dfrac{{\mu \times (5i)}}{{2\pi \times (6 + x)}} = \dfrac{{\mu \times (2i)}}{{2\pi \times (x)}}\]
Simplify
\[(5i)(x) = (2i)(6 + x)\]
\[\Rightarrow 5x = 2x + 12\]
\[\therefore x = 4\,cm\]

So the correct answer is option A.

Note: The direction of the magnetic force on a moving charge is perpendicular to the plane formed by $v$ and $B$ follows right hand rule. The magnitude of the force is proportional to $q, v, B$ and the sine of the angle between $v$ and $B$. A magnetic field is imaginary lines, it only experiences.