Search any question & find its solution
Question:
Answered & Verified by Expert
A solenoid of length $0.4 \mathrm{~m}$ and having 500 turns of wire carries a current $3 \mathrm{~A}$. A thin coil having 10 turns of wire and radius $0.1 \mathrm{~m}$ carries current $0.4 \mathrm{~A}$. the torque required to hold the coil in the middle of the solenoid with its axis perpendicular to the axis of the solenoid is $\left(\mu_0=4 \pi \times 10^{-7}\right.$ SI units, $\left.\pi^2=10\right)\left(\sin 90^{\circ}=1\right)$
Options:
Solution:
1848 Upvotes
Verified Answer
The correct answer is:
$6 \times 10^{-4} \mathrm{Nm}$
The equation for torque is given by:
$$
\tau=\text { NIAB } \sin \theta
$$
As axis of coil and solenoid are perpendicular to each other, $\theta=90^{\circ}$ and $\sin \theta=1$
$$
\begin{aligned}
\therefore \quad \tau & =\mu_0 \mathrm{nI}_1 \times \mathrm{NIA} \\
& =4 \pi \times 10^{-7} \times \frac{500}{0.4} \times 3 \times 10 \times 0.4 \times \pi \times 10^{-2} \\
\tau & =6 \times 10^{-4} \mathrm{Nm}
\end{aligned}
$$
[Note: The answer of the question is not mentioned as an option.]
$$
\tau=\text { NIAB } \sin \theta
$$
As axis of coil and solenoid are perpendicular to each other, $\theta=90^{\circ}$ and $\sin \theta=1$
$$
\begin{aligned}
\therefore \quad \tau & =\mu_0 \mathrm{nI}_1 \times \mathrm{NIA} \\
& =4 \pi \times 10^{-7} \times \frac{500}{0.4} \times 3 \times 10 \times 0.4 \times \pi \times 10^{-2} \\
\tau & =6 \times 10^{-4} \mathrm{Nm}
\end{aligned}
$$
[Note: The answer of the question is not mentioned as an option.]
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.