Question

A current $I=5$A floes along a thin sire as shown in figure, The radius of the curved path of the wire is equal to $R=120$mm. The magnetic induction of the field at the point Q is?

• A. $2.8\times {10}^{-5}T$
• B. $4\times {10}^{-5}T$
• C. $3\times {10}^{-5}T$
• D. $2\times {10}^{-5}$T

Answer 1

The correct option is B $4\times {10}^{-5}T$

Angle covered by ACB part

${360}^{\circ }-{90}^{\circ }={270}^{\circ }$

magnetic field due to ACB part $=\frac{{\mu }_{0}I}{2R}\ast \left(\frac{{270}^{\circ }}{{360}^{\circ }}\right)$

$=\frac{{\mu }_{0}I}{8R}=B_1$

now using formula for straight wire,

$B=\frac{{\mu }_{0}I}{4\pi d}(sin\alpha +sin\beta )$

Here both $\alpha$ and $\beta ={45}^{\circ }$

$d=Rcos{45}^{\circ }=\frac{R}{\sqrt{2}}$

$\therefore B_2=\frac{{\mu }_{0}I}{4\pi \frac{R}{\sqrt{2}}}(sin{45}^{\circ }+sin{45}^{\circ })$

$=\frac{{\mu }_{0}I}{4\pi \frac{R}{\sqrt{2}}}\sqrt{2}=\frac{{\mu }_{0}I}{2\pi R}$

Direction of both magnetic field is same

$\therefore B_{total}=\frac{3{\mu }_{0}I}{8R}+\frac{{\mu }_{0}I}{2\pi R}$

$=\frac{{\mu }_{0}I}{R}[\frac{3}{8}+\frac{1}{2\pi }]$

$=\frac{4\pi \times {10}^{-7}\times 5}{120\ast {10}^{-3}}[0.6+0.16]$

$=0.4\times {10}^{-4}T$

$=4\times {10}^{-5}T$