A magnetic needle lying parallel to a magneti | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The work done $W$ in rotating a magnetic needle from an angle $	heta_1$ to $	heta_2$ in a magnetic field $B$ is given by $W = MB(\cos 	heta_1 -

Vedclass · Accepted Answer

$\sqrt{3} W$

Vedclass · Answer

(A) The work done $W$ in rotating a magnetic needle from an angle $	heta_1$ to $	heta_2$ in a magnetic field $B$ is given by $W = MB(\cos 	heta_1 - \cos 	heta_2)$.Given $	heta_1 = 0^{\circ}$ and $	heta_2 = 60^{\circ}$,we have:$W = MB(\cos 0^{\circ} - \cos 60^{\circ}) = MB(1 - 0.5) = 0.5 MB = \frac{MB}{2}$.Thus,$MB = 2W$.The torque $	au$ required to maintain the needle at an angle $	heta = 60^{\circ}$ is given by $	au = MB \sin 	heta$.Substituting the values,$	au = MB \sin 60^{\circ} = (2W) 	imes \frac{\sqrt{3}}{2} = \sqrt{3} W$.

$A$ magnetic needle lying parallel to a magnetic field requires $W$ units of work to turn it through $60^{\circ}$. The torque required to maintain the needle in this position will be

Similar Questions

Which of the following carries glucose from the digestive tract to the liver?

The energy of the electron in the first orbit of $He^{+}$ is $-871.6 \times 10^{-20} \ J$. The energy of the electron in the first orbit of hydrogen would be

Which salt is used as a purgative?

Which of the following plants exhibits a haplontic life cycle?

When the momentum of a proton is changed by an amount $P_0$,the corresponding change in the de-Broglie wavelength is found to be $0.25\%$. Then,the original momentum of the proton was

Vedclass Test Series

Exam Paper Generator

Online Exam Module