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Nov 24, 2024

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Figure is a graph of the magnetic flux through a certain coil of wire as a function of time during an interval while the radius of the coil s increased, the coi is rotated through 1.5 revolutions, and the external source of the magnetic field s turned off, in that order. Rank the emf induced in the coil at the instants marked A through E from the largest positive value to the largest-magnitude negative value. In your ranking, note any cases of equality and also any instants when the emfs zero.
Solution Q) veriedy Toppr The rankingis B > A > B = D = 0 > C. The emfis given by the negative of the time derivative of the magnetic flux. We pick out the steepest downward slope at instant E as marking the moment of largest emf. Next comes A. At B and at D the graph line is horizontal so the emf is zero. At C' the emf has its greatest negative value
acircular coil of radius 0.1m , 500 turns and resistance 20hm is placed with its plane perpendicular to uniform magnetic field 3T. if its coil is rotated through its vertical diameter through 180deg and 0.25 sec. calculate the magnitude of EMF and current induced in the coil.
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Solution Here the magnetic flux through the circular coil at any instant = BA cosf) where 0/t=0 => 47 (given, t = 0255, 0 = 180°) Therefore induced emfis E= for N turns of coil , E=NBAwsin(ot) E=Eosin(wt); Where Eo is the maximum induced emf. Here A=mr®=71x0.1x0.1=0.017t m? Maximum induced emf Eo=NBA® =500%3.0x0.01xmx47t= 502.65 V Induced current, I = E/r 9265096324
A coil of area 500cm? having 1000 turns is placed such that the plane of the coil is perpendicular to a magnetic field of magnitude 410-weber/m?. If it is rotated by 180 degree about an axis passing through one of its diameter in 0.1 sec, find the average induced emf. A zero. B 3omV ¢ zomV D somV
Solution Q) verinedy Toppr Area of the coil A = 500 cm? = 0.05 m? Number of coil N = 1000 Magnetic field B = 4 x 1075 Wh/m? Magnetic flux passing through the coil initially (0 = 0°), ¢; = NBAcos ¢ = 1000 x 4 x 1075 x 0.05 x ¢0s 0° = 0.002 Wb Magnetic flux passing through the coil finally (8 = 180°), ¢ = NBAcos ¢ = 1000 x 4 x 107 x 0.05 x cos 180° = -0.002 Wb Thus change of flux A¢ = ¢; - ¢ = 0.002 - (-0.002) = 0.004 Wb Change in time At = 0.1s A¢ _0.004_ Thus average emf induced E = 2 40 mV
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Coil of 800 turns of effective area 0. 05 m? is kept perpendicular to a magnetic field of magnitude 5 x 105 T. When the plane of the coil is rotated by 9o°around any of its coplanar axis in 0. 15, the emf induced in the coil will be A 2V c 2x103V D 0.02V
Solution The correct optionis D 0.02 V. Given N = 800, A = 0.05m?, B = 5x105T At = 0.1s (@-0;) a (NBAcos(90°) BAcos(0°) ) at (0-NBA) ar 800x5x10"5x5x10°2 o1 =0.02V
The coil of area 0.1 m? has 500 turns. After placing the coil in a magnetic field of strength 4 x 10~*Wb/m?, if rotated through 90 in 0.1, the average emfinduced in the coil is A 0012V B 005V ¢ o1V
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Solution The correct option is D 0.2 V' o = ~NBAost,costy) = TR 00% 4% 104 0.1(c0890—080) N5 00174 SR 05\ cofoomciio)IIONS o4 o1
Coil of 800 turns of effective area 0. 05 m? is kept perpendicular to a magnetic field of magnitude 5 x 105 T. When the plane of the coil is rotated by 9o°around any of its coplanar axis in 0. 15, the emf induced in the coil will be A 2V c 2x103V D 0.02V
Solution The correct optionis D 0.02 V. Given N = 800, A = 0.05m?, B = 5x105T At = 0.1s (@-0;) e= — a (NBAcos(90°) - NBAcos(0°) )
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