Q. 134.0( 7 Votes )
It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet. A small flat search coil of area 2 cm2 with 25 closely wound turns, is positioned normal to the field direction, and then quickly snatched out of the field region.
Equivalently, one can give it a quick 90° turn to bring its plane parallel to the field direction). The total charge flown in the coil (measured by a ballistic galvanometer connected to coil) is 7.5 mC. The combined resistance of the coil and the galvanometer is 0.50. Estimate the field strength of magnet.
Given: Area of small flat search coil = 2 cm2
Total number of turns in the coil n = 25
Total charge measured by the ballistic galvanometer = 7.5 mC=7.5 ×10-3C
combined resistance of the coil and the galvanometer R= 0.50
The current induced in the coil can be calculated as follows:
i = e/R
substituting in the equation we get
Initial flux through the coil, Ф1 = AB
Final flux through the coil, Ф2= 0
Integrating equation (i) on both sides, we get
⇒ Q = (NBA)/R
Therefore, B = (QR)/NA
So, substituting all the values in above equation, We get
On solving, we get
B = 0.75 T
Rate this question :
How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
PREVIOUSA square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s–1 in the positive x - direction in an environment containing a magnetic field in the positive z - direction. The field is neither uniform in space nor constant in time. It has a gradient of 10–3 T cm–1 along the negative x - direction (that is it increases by 10–3 T cm–1 as one moves in the negative x - direction), and it is decreasing in time at the rate of 10–3 T s–1. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mW.NEXTFigure 6.20 shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. The rails, the rod, and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod = 15 cm, B = 0.50 T, resistance of the closed loop containing the rod = 9.0 mW. Assume the field to be uniform.(a) Suppose K is open and the rod is moved with a speed of 12 cm s–1 in the direction shown. Give the polarity and magnitude of the induced emf.(b) Is there an excess charge built up at the ends of the rods when K is open? What if K is closed?(c) With K open and the rod moving uniformly, there is no net force on the electrons in the rod PQ even though they do experience magnetic force due to the motion of the rod. Explain.(d) What is the retarding force on the rod when K is closed?(e) How much power is required (by an external agent) to keep the rod moving at the same speed (=12cm s–1) when K is closed? How much power is required when K is open?(f) How much power is dissipated as heat in the closed circuit? What is the source of this power?(g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular?
RELATED QUESTIONS :
An air - cored solenoid with length 30 cm, area of cross - section 25 cm2 and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10–3 s. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.NCERT - Physics Part-I
Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity, v = 10 m/s.
Calculate the induced emf in the loop at the instant when x = 0.2 m.
Take a = 0.1 m and assume that the loop has a large resistance.NCERT - Physics Part-I
Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig. 6.21.
NCERT - Physics Part-I
A rectangular metallic loop of length ℓ and width b is placed
coplanarly with a long wire carrying a current i figure. The loop is moved perpendicular to the wire with a speed v in the plane containing the wire and the loop. Calculate the emf induced in the loop when the rear end of the loop is at a distance a from the wire. Solve by using Faraday’s law for the flux through the loop and also by replacing different segments with equivalent batteries.
HC Verma - Concepts of Physics Part 2