Question

A long straight wire carrying a current of $100A$ is placed perpendicular to a uniform magnetic field of strength $1.0\times {10}^{-4}Tesla$ as shown. The resultant field strength at $P$ in magnitude at distance $10cm$ from $O$ is

• A. $2.0\times {10}^{-4}T$
• B. $2.23\times {10}^{-4}T$
• C. $1.73\times {10}^{-4}T$
• D. $0$

Answer 1

The correct option is B $2.23\times {10}^{-4}T$

The magnetic field $B$, due to a long infinite straight current carrying ($I$) wire at a distance $r$ is given by:

$B=\frac{{\mu }_0}{2\pi }.\frac{I}{r}$

Given: $I=100A,r=10cm=0.1m$

Hence, $B=\frac{4\pi \times {10}^{-7}}{2\pi }.\frac{100}{0.1}=2\times {10}^{-4}T$

Now, according to Right hand thumb rule, the direction of this magnetic field at point P will be inwards to the plane of paper. A magnetic field of $1.0\times {10}^{-4}T$ is already there and both the fields are perpendicular to each other.

Therefore resultant field strength at point P will be:

$B^{\prime }=\sqrt{(2\times {10}^{-4}{)}^2+(1.0\times {10}^{-4}{)}^2}=2.23\times {10}^{-4}T$