SR. MPC - 24 - 25 - Magnetic Field due to a Current - 26-8-24.pdf

VINEX TEAM IIT - JEE & NEET ACADEMY SR. MPC - 25...

VINEX TEAM IIT - JEE & NEET ACADEMY SR. MPC - 25 PHY: MAGNETIC FIELDS DURATION: 1 Hr DT: 26-8-24 MAX. MARKS: 100 Marking Scheme: (A) rotate about an axis parallel to the wire (i) Each question is allotted 4 (four) marks for each (B) move away from the wire correct response. (C) move towards the wire (ii) ¼ (one fourth) marks will be deducted for indicating (D) remain stationary. incorrect response of each question. No deduction Q.5 Three particles are moving perpendicular to a from the total score will be made if no response is uniform magnetic field and travel on circular paths indicated for an item in the answer sheet. (figure). They have the same mass and speed. List the –––––––––––––––––––––––––––––––––––––––––––––––– particles in order of their charge magnitude, largest to Q.1 A particle of charge q and mass m moves in a circular smallest. orbit of radius r with angular speed . The ratio of B (out of screen) the magnitude of its magnetic moment to that of its 1 2 angular momentum depends on – (A)  and q (B)  , q and m (C) q and m (D)  and m Q.2 A uniform magnetic field of magnitude 1T exists in region y  0 is along k̂ direction as shown. A particle of charge 1C is projected from point ( 3, 1) towards origin with speed 1 m/sec. If mass 3 of particle is 1 kg, then co-ordinates of centre of circle in which particle moves are – (A) 3, 2, 1 (B) 3, 1, 2 y (C) 2, 3, 1 (D) 1, 2, 3 B=1T Q.6 A flat coil carrying current has a magnetic moment    pm. It is placed in a magnetic field B such that pm  x is anti-parallel to B. The coil is – (A) in stable equilibrium (B) in unstable equilibrium (– 3–1) (C) in neutral equilibrium (D) not in equilibrium (A) (1, 3) (B) (1,  3) q 1 3  3 1 Q.7 If a charged particle of charge to mass ratio   is (C)  ,  (D)  ,  m  2 2   2 2 entering in a magnetic field of strength B at a speed Q.3 An electron with velocity V along the axis v = (2d) (B), then which of the following is correct - approaches a circular current carrying loop as shown 4d in the figure. The magnitude of magnetic force on ×××××××× ×××××××× electron at this instant is – ×××××××× ×××××××× i ×××××××× q/m= ×××××××× ×××××××× V ×××××××× –e ×××××××× ×××××××× ×××××××× x R ×××××××× ×××××××× ×××××××× ×××××××× (A) angle subtended by charged particle at the centre  0 eviR 2 x eviR 2 x (A) (B)  0 of circular path is 2. 2 (x 2  R 2 )3/2 (x 2  R 2 )3/2 (B) the charge will move on a circular path and will come out from magnetic field at a distance 4d  0 eviR 2 x (C) (D) zero from the point of insertion. 4 (x 2  R 2 )3/2 (C) the time for which particle will be in the Q.4 A regular loop carrying a current i is situated near a magnetic field is 2/b. long straight wire such that the wire is parallel to one (D) the charged particle will subtend an angle of 90° of the sides of the loop and is in the plane of the loop. at the centre of circular path If a steady current I established in the wire as shown Q.8 The negatively and uniformly charged nonconducting in the figure, the loop will : disc as shown is rotated clockwise. The direction of i the magnetic field at point A in the plane of the disc is I 1 www.myvinex.com #48-8-1, Somayaji Building, Behind Boy London, Dwarakanagar, Vizag - 16. Ph. No. 0891-2791076, Cell: 9849498582 (1) www.facebook.com/vinex coaching  VINEX TEAM IIT - JEE & NEET ACADEMY (A) evr/2 (B) evr (C) er/2v (D) none of these. Q.15 Which one or more of the following choices is (are) true regarding how a magnetic field acts on a charged A particle. A magnetic field – 1. can increase the speed of a moving particle. 2. can cause a moving particle to accelerate. 3. always exerts a force on a moving particle. (A) into the page (B) out of the page 4. can exert a force on a stationary particle. (C) up the page (D) down the page (A) 2 and 3 (B) 1 and 2 Q.9 At a location near the equator, the earth's magnetic (C) 2 (D) 3 field is horizontal and points north. An electron is Q.16 Four views of a horseshoe magnet and a current- moving vertically upward from the ground. What is carrying wire are shown in the figure. The wire is the direction of the magnetic force that acts on the perpendicular to the screen, and the current is electron? directed out of the screen toward you. In which one (A) North (B) East or more of these situations does the magnetic force (C) South (D) West on the current point due north? Q.10 A uniform magnetic field exists in region which Current (out of screen) forms equilateral triangle of side a. The magnetic N field is perpendicular to the plane of the triangle. A charge q enters into this magnetic field perpendicularly with speed v along perpendicular bisector of one side and comes out along perpendicular bisector of other side. The magnetic 1 2 3 4 S induction in the triangle is – mv 2mv (A) 1 and 2 (B) 3 and 4 (A) (B) (C) 2 (D) 1 qa qa Q.17 A long straight wire, carrying current I, is bent at its mv mv midpoint to form an angle of 45°. Magnetic field at (C) (D) 2qa 4qa point P, distance R from point of bending is equal to  P I Q.11 A particle is moving with velocity v  ˆi  3jˆ and it  R 45° produces an electric field at a point given by E  2kˆ. I It will produce magnetic field at that point equal to (all quantities are in SI units) 6iˆ  2jˆ ( 2  1)  0 I ( 2  1) 0 I (A) (A) (B) c2 4 R 4 R 6iˆ  2jˆ ( 2  1) 0 I ( 2  1)  0 I (B) (C) (D) c2 4 2 R 2 2 R (C) zero Q.18 A uniform magnetic field of 1.5 T exists in a (D) cannot be determined from the given data cylindrical region of radius 10.0 cm, it’s direction Q.12 A charged particle P leaves the origin with speed being parallel to the axis along east to west. A current v = v0, at some inclination with the x-axis. There is carrying wire in north south direction passes through uniform magnetic field B along the x-axis. P strikes a this region. The wire intersects the axis and fixed target T on the x-axis for a minimum value of experience a force of 1.2 N downward. B = B0. P will also strike T if – If the wire is turned from North south to north east- (A) B = 2B0, v = 2v0 (B) B = 2B0, v = v0 south west direction, then magnitude and direction of (C) B = B0, v=2v0 (D) B= B0/2, v = 2v0. force is – Q.13 A current (I) carrying circular wire of radius R is (A) 1.2 N, upward (B) 1.2 2 , downward placed in a magnetic field B perpendicular to its 1.2 (C) 1.2 N, downward (D) N , downward plane. The tension T along the circumference of the 2 wire is Q.19 A long straight wire along the z-axis carries a current (A) BIR (B) 2BIR  I in the negative z direction. The magnetic vector B (C) BIR (D) 2BIR field at a point having coordinates (x, y) in the z = 0 Q.14 The magnetic moment of an electron orbiting in a plane is – circular orbit of radius r with a speed v is equal to: www.myvinex.com #48-8-1, Somayaji Building, Behind Boy London, Dwarakanagar, Vizag - 16. Ph. No. 0891-2791076, Cell: 9849498582 (2) www.facebook.com/vinex coaching  VINEX TEAM IIT - JEE & NEET ACADEMY  0 I (yiˆ  xj) ˆ  0 I (xiˆ  yj) ˆ Q.23 An electron with mass m, velocity v and charge e (A) (B) describes half a revolution in a circle of radius r in a 2 (x 2  y2 ) 2 (x 2  y2 ) magnetic field B, will acquire energy equal to  0 I (xjˆ  yi) ˆ  0 I (xiˆ  yj) ˆ Q.24 An infinitely long conductor PQR is bent to form a (C) (D) right angle as shown. A current I flows through PQR. 2 (x 2  y2 ) 2 (x 2  y2 ) The magnetic field due to the current at the point M Q.20 An infinitely long wire carrying current I is along Y is H1. Now, another infinitely long straight conductor axis such that its one end is at point A (0, b) while the QS is connected at Q so that the current is I/2 in QR wire extends upto +  The magnitude of magnetic as well as in QS, the current in PQ remaining field strength at point (a, 0). unchanged. The magnetic field at M is now H2. The ratio H1/H2 is given by 2 : X. Find the value of X. M I A I 90° (0,0) (a,0) – + P Q 90° S 0 I  b  0 I  b  (A) 1   (B) 1   R 4 a  a 2  b2  4 a  a2  b2  0 I  b  (C) 1   (D) None of these – 4 a  a2  b2  Q.25 The rectangular coil having 100 turns is turned in a For Q.21-Q.25 : 0.05 uniform magnetic field of ĵ Tesla as shown in The answer to each question is a NUMERICAL 2 VALUE. the figure. The torque acting on the loop is Q.21 The radii of two concentric coils having same number of turns are 10 cm and 20 cm respectively. 5.66 × 10–X (N-m) k̂. Find the value of X.. z Equal currents are passed through them first in same direction and then in opposite direction. In these two conditions, The ratio of resultant magnetic fields at 0.08m the centre is Q.22 A current loop is shown edge-on in the figure. The I=0.5A loop carries 10 amps and has a surface area of 0.001 y m2 in a magnetic field of 2 T; the torque on it -0.04m (oriented as shown) is 1 Nm. What is the number of turns? x I out I 30 B I in www.myvinex.com #48-8-1, Somayaji Building, Behind Boy London, Dwarakanagar, Vizag - 16. Ph. No. 0891-2791076, Cell: 9849498582 (3) www.facebook.com/vinex coaching