Question

proton -18 V.7.20 8 + 1.50x10"" -0.05CC X IN our straight wive left word current 17.1 A wire king the date given by the diagram (a) Foud' the diction of the wire's maguetre field wire at the location occupied by Auscel, circle oue below B field direction is (5) Calculate the strength of B at the location of f Aus(s) 13 wivel 6) Celculate the magnitude of the magnetic Sover F on the moving charge 8. Ans le FL (d) Detenaive the direction of the magnetic Sorce on q Ans (of cinde one below ? F direction is In our A rectangular coil consisting of so rectangular loops 0.050m In length by 0.030m i widty is held broadside to a magnet's strong maquete field of 8.0x102 Tesle. Over a time interval of 3.4 xio?s the magret os pulled away, so that the 13- Sield perpendicular to the cuil is reduced 6 2.0 x10 *T. What is the average erf riduced the coll over this time lutervel? 동 final (+- 3.4 x10"15) 50 loep Coil zútil (t=-) Field lines pass perpendicularly through coil coil skape 50 turns 0.050ml w 0.030 M Area A la : Ans Ecoil

Answer 1

magnetic field due to long straight wire carrying current i, at a point is given by Moi B= 2 x ar ris the distance of the point from wire The direction of the magnetic field can be determined by using right hand thumb rule. Point the thumb of right hand in the current's direction then curl the fingers around the wire which points in the direction of magnetic field. proton i right According to right hand rule stated above the direction of magnetic field due to the wire at the location of the proton is into the page in magnitude of the magnetic field B= Mol 2 xr * 10-7)(17) 4x7x 10-7 B= (2x) * (0.055) B = 6.182 x 10-5T Bwire = 6.182 x 10-57 direction into the page force on a positive charge particle(+9) proton moving in a magnetic field (B) with

velocity (v) is given by ух в This force is in the direction of V B for a positive charge (+q) proton Now the direction of х B can be determined according right hand cross product rule If fingers of right hand are pointed in the direction of velocity vector v and curved in the direction of magnetic field vector B then thumb points in the direction of VX B Left into the page Å force Therefore the direction of the magnetic force is downward Magnitude of the magnetic force F = qBv

F= = (1.6 x 10-19)> (6.182 x 10+5)* (7* 100) * 1017 F = 6.924 x 10-17 Direction as explained above is towards downward → 4 According to faraday's law average EMF induced is 40 E = 0 = NBAcos A = 0 A = lw Ø = NBlw AB E = - Nlw BPB; E = - Nlw t E = -(50) (0.05) (0.03) * 2 x 10-2-8 10-2 3.4 x 10-3 1.324 Ecoil 1.324V