Magnetic Effects Of Current Question 238
Question: As shown in the figure, a three-sided frame is pivoted at P and Q and hangs vertically. Its sides are of same length and have a linear density of $ \sqrt{3} $ kg/m. A current of $ 10\sqrt{3} $ A is sent through the frame, which is in a uniform magnetic field of 2T directed upwards as shown. Then angle through which the frame will be deflected in equilibrium is $ (Take,g=10,m/s^{2}) $
Options:
A) $ 30^{0} $
B) $ 45^{0} $
C) $ 60^{0} $
D) $ 90^{0} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Torque of magnetic force about PQ $ {\tau_{m}}=(ILB)Lcos\theta =IL^{2}B\cos \theta $ Torque of gravitational force about PQ $ {\tau_{g}}=[(\lambda L)gLsin\theta +2(\lambda L)g(1/2)Lsin\theta ]\lambda L^{2}g\sin \theta $ $ {\tau_{m}}={\tau_{g}}\Rightarrow \tan \theta =\frac{IB}{2\lambda g}=\frac{10\sqrt{3}\times 2}{2\times \sqrt{3}\times 10}=1 $
$ \Rightarrow \theta =45{}^\circ $
BETA
