Question

A current of $ 50 A$is passed through a straight wire of length $ 6 cm$, then the magnetic induction at a point$ 5 cm$ from either end of the wire is? ($ 1 gauss={10}^{-4}T$)

B

• A. $2.5gauss$
• B. $1.25gauss$
• C. $1.5gauss$
• D. $3.0gauss$

Answer 1

The correct option is C

$1.5gauss$

Step 1: Given data

Electric current flowing through the straight conductor, $i=50A$

Length of the wire, $L=6cm$

Distance of point ($p$) from the centre of the wire, $r=4cm=0.4m$

$1gauss={10}^{-4}T$

Angle between the point P and the one end of the wire,${\theta }_1={37}^0$

Angle between the point P and the other end of the wire,${\theta }_2={37}^0$

Magnetic permeability, ${\mu }_0=4\pi \times {10}^{-7}$

Step 2: Formula used

$B=\frac{{\mu }_{0}i}{4\pi r}\left(\sin {\theta }_1+\sin {\theta }_2\right)$ …………………….(a)

$B$ denotes the magnetic induction at the point P

Step 3: Calculation of the magnetic induction at point P

Substituting the given values in equation (a), we get

$$\begin{gathered} B=\frac{4\pi \times {10}^{-7}\times 50A}{4\pi \times 0.04m}\left(\sin {37}^0+\sin {37}^0\right) \\ B=\frac{{10}^{-7}\times 50}{0.04}(0.6+0.6) \\ B=1.5\times {10}^{-4}T \\ B=1.5gauss \end{gathered}$$

Hence, option C is correct.