MHT CET · Physics · Magnetic Effects of Current
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, \(\left.\pi^2=10\right)\left(\sin 90^{\circ}=1\right)\)
- A \(3 \times 10^{-4} \mathrm{Nm}\)
- B \(12 \times 10^{-4} \mathrm{Nm}\)
- C \(6 \times 10^{-4} \mathrm{Nm}\)
- D \(24 \times 10^{-4} \mathrm{Nm}\)
Answer & Solution
Correct Answer
(C) \(6 \times 10^{-4} \mathrm{Nm}\)
Step-by-step Solution
Detailed explanation
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 \(\sin \theta=1\)
\(\therefore \tau =\mu_0 \mathrm{nI}_1 \times \mathrm{NIA}\)
\(=4 \pi \times 10^{-7} \times \frac{500}{0.4} \times 3 \times 10 \times 0.4 \times \pi \times 10^{-2}\)
\(\tau =6 \times 10^{-4} \mathrm{Nm}\)
[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 \(\sin \theta=1\)
\(\therefore \tau =\mu_0 \mathrm{nI}_1 \times \mathrm{NIA}\)
\(=4 \pi \times 10^{-7} \times \frac{500}{0.4} \times 3 \times 10 \times 0.4 \times \pi \times 10^{-2}\)
\(\tau =6 \times 10^{-4} \mathrm{Nm}\)
[Note: The answer of the question is not mentioned as an option.]
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