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 cm^{2} 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.

Answer :

Given: Area of small flat search coil = 2 cm^{2}

Total number of turns in the coil n = 25

Total charge measured by the ballistic galvanometer = 7.5 mC=7.5 ×10^{-3}C

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}

_{also,}

substituting in the equation we get

⇒ …………..(i)

Initial flux through the coil, Ф_{1} = AB

Final flux through the coil, Ф_{2}= 0

Integrating equation (i) on both sides, we get

Total charge

Therefore,

⇒ 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

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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?

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