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Physics NEET MCQ
Magnetic Effects Of Current Mcq
Quiz 12
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Q.1
An electron is moving in a circular path under theinfluence of a transverse magnetic field of 3.57 × 10–2 T. If the value of e/m is 1.76 × 1011 C/kg, the frequency of revolution ofthe electron is…[NEET II-2016]
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a) 62.8 MHz
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b) 6.28 MHz
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c) 1 GHz
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d) 100 MHz
Explanation
Magnetic force on moving charge f=evB Centripetal force v=ωr Answer:(c)
Q.2
A 250-Turn rectangular coil of length 2.1 cm andwidth 1.25 cm carries a current of 85 μA and subjected to a magnetic field of strength 0.85 T.Work done for rotating the coil by 180° against the torque is [ NEET 207]
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a) 9.1 µJ
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b) 4.55 µJ
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c) 2.3 µJ
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d) 1.15 µJ
Explanation
W = MB [cosθ1 – cosθ2] As θ1 = 0 and θ2 = 180° W = 2MB Now M = nIA W= 2×250×85×10-6×( 2.1×1.25×10-4)×0.85= 9.48×10-6 W = 9.1 µJ Answer:(a)
Q.3
An arrangement of three parallel straight wiresplaced perpendicular to plane of paper carrying samecurrent ‘I’ along the same direction is shown in Fig.Magnitude of force per unit length on the middle wire‘B’ is given by…[ NEET 2017]
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a)
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b)
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c)
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d)
Explanation
Current magnitude and direction is same, force are mutually perpendicular FA = FC =F F'=√2 F Answer:(c)
Q.4
A particle of mass M and positive charge Q, moving with a constant velocity u1 =4(i ) m/s2, enters a region of uniform static magnetic field normal to the x-y plane.The region of the magnetic field extends from x = 0 to x = L for all values of y. Afterpassing through this region, the particle emerges on the other side after 10 millisecondswith a velocity The correct statement(s) is (are)
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a) The direction of the magnetic field is −z direction.
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b) The direction of the magnetic field is +z direction.
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c) The magnitude of the magnetic field 50πM/3Q units.
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d) The magnitude of the magnetic field is 100πM/3Q units.
Explanation
Since final velocity x component decreased and y-component increased charge is moving in anti-clock wise direction magnetic filed in in –Z axis Calculation of v charge moved from a to b in 10 milisec v=ab/t But ab= rθ ∴v=rθ/t Substituting value of v in (i) From figure and we get tanθ =1/√3 ⇒ θ= π/6 On substituting value of θ and t in (ii) we get Answer:(a, c)
Q.5
A steady current I flows along an infinitely long hollow cylindrical conductor of radius R.This cylinder is placed coaxially inside an infinite solenoid of radius 2R. The solenoid hasn turns per unit length and carries a steady current I. Consider a point P at a distance rfrom the common axis. The correct statement(s) is (are)…[ IIT advance 2013]
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a) In the region 0 < r < R, the magnetic field is non-zero.
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b) In the region R < r < 2R, the magnetic field is along the common axis.
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c) In the region R < r < 2R, the magnetic field is tangential to the circle of radius r,centered on the axis.
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d) In the region r > 2R, the magnetic field is non-zero.
Explanation
Magnetic field produced by hollow cylindrical conductor on outside is similar to thin conductor and is perpendicular to plane of paper. No magnetic field will be produced inside the cylinder. Solenoid will produce magnetic field in side but not outside parallel to plane of paper Option a ; correct as magnetic field inside the cylinder is due to solenoid Option b: Wrong as resultant magnetic filed will produce , and will not be along common axis Option c: Wrong ,Explanation same as b Option d: correct, magnetic filed will be due to cylinder Answer:(a, d)
Q.6
Paragraph The figure shows a circular loop of radius a withtwo long parallel wires (numbered 1 and 2) all inthe plane of the paper. The distance of each wirefrom the centre of the loop is d. The loop and thewires are carrying the same current I. The currentin the loop is in the counterclockwise direction ifseen from above. Q338A) When d ≈ a but wires are not touching the loop, it is found that the net magnetic field onthe axis of the loop is zero at a height h above the loop. In that case
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a) current in wire 1 and wire 2 is the direction PQ and RS, respectively and h ≈ a
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b) current in wire 1 and wire 2 is the direction PQ and SR, respectively and h ≈ a
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c) current in wire 1 and wire 2 is the direction PQ and SR, respectively and h ≈ 1.2 a
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d) current in wire 1 and wire 2 is the direction PQ and RS, respectively and h ≈ 1.2 a
Explanation
Magnetic filed due to wires at axis of loop above = Magnetic filed at above h from centre of loop Magnetic filed due to wire at the axis of loop at distance h from centre of loop We know that direction of magnetic filed at point is tangential to that point Thus We have to consider cos component of BW( magnetic filed due to wire) Magnetic filed due to loop at height h Magnetic filed due to two wires = Magnetic field due to loop h ≈1.21a Answer:(c)
Q.7
Consider d >> a, and the loop is rotated about its diameter parallel to the wires by 30° from the position shown in the figure. If the currents in the wires are in the oppositedirections, the torque on the loop at its new position will be (assume that the net field dueto the wires is constant over the loop)
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a)
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b)
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c)
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d)
Explanation
τ= MBsinθ M= nIA = Iπa2 Answer:(b)
Q.8
Four charges Q1, Q2, Q3 and Q4 of same magnitude are fixed along the x axis at x = −2a,−a, +a and +2a, respectively. A positive charge q is placed on the positive y axis at adistance b >Four options of the signs of these charges are given in List I. The directionof the forces on the charge q is given in List II. Match List I with List II and select thecorrect answer using the code given below the lists. Code :
List I
List II
P. Q
1
, Q
2
, Q
3
, Q
4
all positive
+x
Q. Q
1
, Q
2
positive; Q
3
, Q
4
negative
−x
R. Q
1
, Q
4
positive; Q
2
, Q
3
negative
+y
S. Q
1
, Q
3
positive; Q
2
, Q
4
negative
−y
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a) P-3, Q-1, R-4, S-2
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b) P-4, Q-2, R-3, S-1
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c) P-3, Q-1, R-2, S-4
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d) P-4, Q-2, R-1, S-3
Explanation
List I P All x component gets cancelled only +y components gets added P → 3 List I Q All y components gets cancelled and only +x components gets added Q →1 List I R Each pair cancel x of each other and y of +ve pair will be smaller than –ve so net will be along –ve y. R → 4 List I S Each dipole pair will cancelled y of each other and of Q2 -Q3 pair along –ve x is stronger than that of Q1 – Q2 along +ve x S→2 P → 3, Q → 1, R → 4, S →2 Answer:(a)
Q.9
A conductor (show in the figure) carrying constant current I is kept in the x-y plane in auniform magnetic field B. If F is the magnitude of the total magnetic force acting on theconductor, then the correct statement(s)is(are) …[IIT Advance 2015]
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d) If B is along z , F = 0
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a) If B is along z
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b) If B is along x , F =0
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c) If B is along y , F ∝ (L + R)
Explanation
Option a B is along z Force of L = Bil Force on R = BiR Total force = 2Bil+ 2BiR F ∝ (L + R) aCorrect and d is wrong Option B , B is along X Force on L = 0 But force of R= 0 Option b correct Option c B along y axis Force on L = Bil Force on R = BiRsinθ Option C correct Answer:(a, b, c)
Q.10
In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in the figure. The length, width and thickness of the strip are l,w and d, respectively. A uniform magnetic field B is applied on the strip along the positive y-direction. Due to this, the charge carriers experience a net deflection along the z-direction. This results in accumulation of charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues until the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross section of the strip and carried by electrons. Q342A) Consider two different metallic strips (1 and 2) of the same material. Their lengths are the same, widths are w1 and w2 and thicknesses are d1 and d2, respectively. Two points K and M are symmetrically located on the opposite faces parallel to the x-y plane (see figure). V1 and V2 are the potential differences between K and M in strips 1 and 2, respectively. Then, for a given current I flowing through them in a given magnetic field strength B, the correct statement(s) is (are)
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a) If w1 = w2 and d1 = 2d2, then V2 = 2V1
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b) If w1 = w2 and d1 = 2d2, then V2 = V1
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c) If w1 = 2w2 and d1 = d2, then V2 = 2V1
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d) If w1 = 2w2 and d1 = d2, then V2 = V1
Explanation
Option a If B1 = B2 and n1 = 2n2 V ∝ (1/n) V1 n1 = V2 n2 2V1 n21 =V2 n2 2V1 = V2 Option a correct , option b wrong Option c : If B1 = 2B2 and n1 = n2, then V2 = 0.5V1 V ∝ B Option c correct. Option d wrong Answer:(a,c)
Q.11
A conducting loop in the shape of a right angled isosceles triangle of height 10 cm is kept such that the 90° vertex is very close to an infinitely long conducting wire (see the figure). The wire is electrically insulated from the loop. The hypotenuse of the triangle is parallel to the wire. The current in the triangular loop is in counterclockwise direction and increased at a constant rate of 10 A s-Which of the following statement(s) is(are) true?
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a) The induced current in the wire is in opposite direction to the current along the hypotenuse
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b) The magnitude of induced emf in the wire is volt (µ0/π) volts
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c) There is a repulsive force between the wire and the loop
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d) If the loop is rotated at a constant angular speed about the wire, an additional emf of (µ0/π) volt is induced in the wire
Explanation
Current in wire will be in the direction of current in hypoteneous as current is decreasing option a wrong Option c is correct Force is repulsive follows from Lenz Law Consider the reciprocal case. Current is flowing through the straight wire and electric potential is induced in conducting triangular loop. Consider a small part of loop at distance r and with dr. Length of segment , from trigonometry 2r Area of segment 2rdr Magnetic field produced is (μ0 I)/2πr Thus flux Now if current is passing through loop then ϕ is flux associated with wire Form the formula for mutual inductance M=Φ/I M=(µ0 l)/π Induced emf As di/dt = 10 A/s and l = 10-1m E=µ0/π Option b correct If loop is rotated with constant angular acceleration additional emf will not be induced . As there is no change in magnetic flux option d wrong Answer:(b, c)
Q.12
Consider two identical galvanometers and two identical resistors with resistance R. If the internal resistance of the galvanometers RC < R/2, which of the following statement(s) about any one of the galvanometers is(are) true? ..[ IIT Advance 2016]
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a) The maximum voltage range is obtained when all the components are connected in series
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b) The maximum voltage range is obtained when the two resistors and one galvanometer are connected in series, and the second galvanometer is connected in parallel to the first galvanometer
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c) The maximum current range is obtained when all the components are connected in parallel
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d) The maximum current range is obtained when the two galvanometers are connected in series and the combination is connected in parallel with both the resistors
Explanation
V = Ig(R+RC) Option a V = Ig 2(R+RC) =Ig(2R+2RC) Option b Galvanometers are parallel, then total current capacity increases to 2Ig and both are in series with parallel combination R’C = RC/2 and 2R Given Rc < R/2 Thus option b have more range For Ammeter Option c Effective resistance of galvanometer and other two is RC/2 and R/2 Combined capacity of current of galvanometer 2Ig Option d Resistance of two galvanometer connected in series = 2RC . Parallel combination of resistance is R/2 Option c have more range than option d as RC < R/2 Answer:(b, c)
Q.13
A, Q.345B and Q.345C by appropriately matching the information given in the three columns of the following table. Q345 )A charged particle (electron or proton) is introduced at the origin (x = 0, y = 0, z = 0) with a given initial velocity v. A uniform electric field (E ) and a uniform magnetic field B exist everywhere. The velocity v , electric field E and magnetic field B are given in column 1, 2 and 3, respectively. The quantities E0, B0 are positive in magnitude Q345A . In which case will the particle move in a straight line with constant velocity ?
column 1
column 2
column 3
(I) Electron with
(i)
E
=E
0
z
(P)
B
= -B
0
x
(II) Electron with
(ii)
E
= -E
0
y
(Q)
B
=B
0
x
(III) Proton with
v
= 0
(iii)
E
= -E
0
x
(R)
B
= B
0
y
(IV) Proton with
(iv)
E
= E
0
x
(S)
B
=B
0
z
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a) (II) (iii) (S)
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b) (IV) (i) (S)
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c) (III) (ii) (R)
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d) (III) (iii) (P)
Explanation
For particle to move in straight line resultant force =0 Ee = e(v×B ) Thus vector multiplication of v and B should be along direction of E Which possible only in option a Answer:(a)
Q.14
A symmetric star shaped conducting wire loop is carrying a steady state current I as shown in the figure. The distance between the diametrically opposite vertices of the star is 4a. The magnitude of the magnetic field at the center of the loop is … [ IIT Advance2017]
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a)
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b)
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c)
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d)
Explanation
From the diagram perpendicular distance of pq = a, Angle qos = 60, thus angle qor =β=30O and por =α =60O From the formula There are 12 such segments Total B Answer:(c)
Q.15
) A uniform magnetic field B exists in the region between x = 0 and x = 3R/2 (region 2 in the figure) pointing normally into the plane of the paper. A particle with charge +Q and momentum p directed along x-axis enters region 2 from region 1 at point P1(y = –R). Which of the following options(s) is/are correct ? [ IIT Advance 2017]
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a) For the particle will enter region 3 through the point P2 on x-axis
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b) For the particle will re-enter region 1
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c) For a fixed B, particle of same charge Q and same velocity v, the distance between the point P1 and the point of re-entry into region 1 is inversely proportional to mass of the particle
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d) When the particle reinters region 1 through the longest possible path in region 2, the magnitude of the change in its linear momentum between point P1 and the farthest point from y-axis is p/√2
Explanation
Centripetal force is magnetic From figure O’P2 = 13R/8 and OP2 = 3R/2. Then OO’ = 5R/8 If radius r < 13R/8 charge will not enter region 3 from point P2 Option a correct Now 2/3 p/QR thus radius = 3R/2 Since 3R/2 < 13R/8 charge will not enter region 3 and will enter region 1 Option b correct For reentry in region 1 maximum radius Thus reentry in region 1 is proportional to mass [ Option c wrong] When charge re-enter region direction of momentum is reversed Thus Δ p=p i - pj |Δp|=p√2 Option d is wrong Answer:(a,b)
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