Question

A long wire carries a current of $18\, A$ kept along the axis of a long solenoid of radius $1\, cm$. The field due to the solenoid is $8.0 \times 10^{-3} T$. The magnitude of the resultant field at a point $0.6\, mm$ from the solenoid axis is (Assume, $\mu_{0}=4 \pi \times 10^{-7} Tm / A$ )

• A. $6 \times 10^{-3} T$
• B. $6 \times 10^{-4} T$
• C. $2 \sqrt{7} \times 10^{-3} T$
• D. $10 \times 10^{-3} T$

Answer 1

Answer: (D)

Given, current in a long wire, $I=18 \,A$

magnetic field due to a solenoid,
$B_{1}=8 \times 10^{-3} T$
and $r=0.6 \,mm =0.6 \times 10^{-3}\,m$
Radius of solenoid, $R=1 \,cm =1 \times 10^{-2} \,m$
Magnetic field due to current carrying wire at a point, $P$
$B_{2} =\frac{\mu_{0}}{2 \pi} \cdot \frac{I}{r}=2 \times 10^{-7} \times \frac{18}{0.6 \times 10^{-3}} $
$=6 \times 10^{-3} T$ (in upward direction)
Since, current carrying wire is placed along the axis of solenoid, hence magnetic field produced by wire and magnetic field due to solenoid at point $P$, both are perpendicular to each other, Hence, the magnitude of the resultant magnetic field,
$B =\sqrt{B_{1}^{2}+B_{2}^{2}}$
$=\sqrt{\left(8 \times 10^{-3}\right)^{2}+\left(6 \times 10^{-3}\right)^{2}} $
$=10 \times 10^{-3} T$