Magnetism Question 38
Question: A small coil C with N = 200 turns is mounted on one end of a balance beam and introduced between the poles of an electromagnet as shown in figure. The cross sectional area of coil is A= 1.0 cm2, length of arm OA of the balance beam is $ l=30,cm. $ When there is no current in the coil the balance is in equilibrium. On passing a current I = 22 mA through the coil the equilibrium is restored by putting the additional counter weight of mass $ \Delta m=60,mg $ on the balance pan. Find the magnetic induction at the spot where coil is located.
Options:
A) 0.4 T
B) 0.3 T
C) 0.2 T
D) 0.1 T
Show Answer
Answer:
Correct Answer: A
Solution:
On passing current through the coil, it acts as a magnetic dipole. Torque acting on magnetic dipole is counter balanced by the moment of additional weight about position O. Torque acting on a magnetic dipole $ \tau =MB\sin \theta =(NiA)B\sin 90^{o}=NiAB $ . Again $ \tau =Force\times Lever\ arm\ =\Delta mg\times l $
$ \Rightarrow NiAB=\Delta mgl $
$ \Rightarrow B=\frac{\Delta mgl}{NiA}=\frac{60\times {{10}^{-3}}\times 9.8\times 30\times {{10}^{-2}}}{200\times 22\times {{10}^{-3}}\times 1\times {{10}^{-4}}} $ = 0.4 T
BETA
