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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Consider six wires coming into or out of the page, all with the same current. Rank the line integral of the magnetic field (from most positive to most negative) taken counterclockwise around each loop shown A. `BgtCgtDgtA`B. `BgtC=DgtA`C. `BltAgtC=D`D. `CltB=DgtA` |
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Answer» Correct Answer - C (c) Loop B:`mu_0(2i-i)=ointB.dl` Loop C:`mu_0(i-2i)=ointB.dl` Loop A:`mu_0(3i-3i)=ointB.dl` Loop D:`mu_0(0-i)=ointB.dl` (c) `BgtAgtC=D` |
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| 2. |
Two positive charges `q_1 and q_2` are moving with velocities `v_1 and v_2` when they are at points A and B, respectively, as shown in Fig. The magnetic force experienced by charge `q_1 ` due to the other charge `q_2` is A. `(mu_0q_1q_2v_1v_2)/(8sqrt2pia^2)`B. `(mu_0q_1q_2v_1v_2)/(4sqrt2pia^2)`C. `(mu_0q_1q_2v_1v_2)/(2sqrt2pia^2)`D. `(mu_0q_1q_2v_1v_2)/(sqrt2pia^2)` |
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Answer» Correct Answer - A (a) Magnetic field at A due to `q_2`: `vecB_2=(mu_0)/(4pi) (q_2v_2rsin 45^@)/(r^3)hatk` where `r=sqrt2a` `implies vecB_2=(mu_0)/(4pi) (q_2v_2)/(2sqrt2a^2) hatk` Force on `q_1 : vecF=q_1vecv_1xxvecB_2=(mu_0)/(4pi) (q_1q_2v_1v_2)/(2sqrt2a^2)(-hatj)` |
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| 3. |
A circular current carrying coil has a radius R. The distance from the centre of the coil on the axis where the magnetic induction will be `(1)/(8)` th to its value at the centre of the coil, isA. `(R)/(sqrt(3))`B. `Rsqrt(3)`C. `2sqrt(3)R`D. `(2)/(sqrt(3))R` |
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Answer» Correct Answer - C |
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| 4. |
An otherwise infinite, straight wire has two concentric loops of radii a and b carrying equal currents in opposite directions as shown in Fig. The magnetic field at the common centre is zero for A. `a/b=(pi-1)/pi`B. `a/b=pi/(pi+1)`C. `a/b=(pi-1)/(pi+1)`D. `a/b=(pi+1)/(pi-1)` |
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Answer» Correct Answer - B (b) `B_(centre)=0` `(mu_0I)/(4pib)odot+(mu_0I)/(4pib)odot +(mu_0I)/(2b)odot +(mu_0I)/(2a)ox=0` `implies (mu_0I)/(2pib)+(mu_0I)/(2b)-(mu_0I)/(2a)=0 implies 1/(2pib)+1/(2b)=1/(2a)` `implies a/b=pi/(pi+1)` |
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| 5. |
`L` is a circular ring made of a uniform wire, currents enters and leaves the ring through straight conductors which, if produces, would have passed through the centre `C` of ring. The magnetic field at `C` (i) due to the straight conductors is zero (ii) due to the loop is zero (iii) due to the loop is proportional to `theta` (iv) due to loop is proportional to `(pi-theta)`A. due to the straight conductors is zeroB. due to the loop is zeroC. due to path II is proportional to `theta`.D. due to the path I is proportional `(pi-theta)` |
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Answer» Correct Answer - B |
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| 6. |
A long insulated copper wire is closely wound as a spiral of `N` turns. The spiral has inner radius a and outer radius `b`. The spiral lies in the `xy`-plane and a steady current I flows through the wire. The`z`-component of the magetic field at the centre of the spiral is A. `(mu_0NI)/(2(b-a)) In (b/a)`B. `(mu_0NI)/(2(b-a)) In ((b+a)/(b-a))`C. `(mu_0NI)/(2b) In (b/a)`D. `(mu_0NI)/(2b) In ((b+a)/(b-a))` |
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Answer» Correct Answer - A (a) `B=int(mu_0(dN)I)/(2x)= int_a^b(mu_0(N/(b-a)dx)I)/(2x)=(mu_0NI)/(2(b-a))ln(b/a)` |
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| 7. |
Four parallel conductors, carrying equal currents, pass vertically through the four corners of a square WXYZ. In two conductors, the current is flowing into the page, and in the other two out into the page, and in the other two out of the page. In what directions must the currents flow to produce a resultant magnetic field in the direction shown at O, the centre of the square? A. W and Y X and ZB. X and Z W and YC. W and Z X and YD. W and X Y and Z |
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Answer» Correct Answer - D (d) To get magentic field in resultant direction, current in X should be in and that in Z should be out, I in W should be in and that in Y should be out. |
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| 8. |
Two protons move parallel to each other with an equal velocity `v=300kms^-1`. Find the ratio of forces of magnetic and electric interaction of the protons. |
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Answer» Correct Answer - `1.00xx10^-6` The force of magnetic induction is given by `vecF_(mag)=e(vecvxxvecB) Here vecB=(mu_0)/(4pi) (e(vecvxxvecr))/(r^3)` `:. vecF_(mag)=(mu_0)/(4pi)xx(e^2)/(r^3)[vecvxx(vecvxxvecr)]` `=(mu_0)/(4pi)xx(e^2)/(r^3)[(vecv.vecr)xxvecv-(vecv.vecv)xxvecr]` `=(mu_0)/(4pi)xx(e^2)/(r^3)xx(-v^2vecr)` `vecF_("ele.")=evecE=e 1/(4piepsilon_0) (evecr)/(r^3)` `:. |F_(mag)|/|F_("ele.")|=v^2mu_0epsilon_0=(v/c)^2=1.00xx10^-6` |
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| 9. |
Figure. Shows a small loop carrying a current I. The curved portion is an arc of a circle of radius R and the straight portion is a chord to the same circle subtending an angle `theta`. The magnetic induction at centor O is A. zeroB. always inward irrespective of the value `theta`C. inward as long as `theta` is less than `pi`D. always outward irrespective of the value of `theta` |
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Answer» Correct Answer - C (c) Due to current shown in figure, the magnetic field with loop is outwards. So field at O should be inward because magnetic field lines from closed loop. But if `theta gt pi`, then point O will lie within loop and for a point within loop magnetic field is outward. |
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| 10. |
A cylindrical cavity of diameter a exists inside a cylinder of diameter `2a` as shown in the figure. Both the cylinder and the cavity are infinitity long. A uniform current density `j` flows along the length . If the magnitude of the magnetic field at the point `P` is given by `(N)/(12) mu_(0)aJ`, then the value of `N` is |
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Answer» Correct Answer - `(5)` (5) `B_1=(mu_Jpia^2)/(2pia)=(mu_0Ja)/2,B_2=(mu_Jpi(a//2)^2)/(2pi(3a//2))=(mu_0Ja)/12` `B_1-B_2=(mu_Ja)/2-(mu_Ja)/12` `=((mu_0Ja)/2)(1-1/6)=5/6((mu_0Ja)/2)=(5mu_0aJ)/12` `=N/12mu_0aJ` `:. N=5` |
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| 11. |
A steady electric current is flowing through a cylindrical conductor. Then,A. the electric field at the axis of the conductor is zeroB. the magnetic field at the axis of the conductor is zeroC. the electric field in the vicinity of the conductor is zeroD. the magnetic field in the vicinity of the conductor is zero |
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Answer» Correct Answer - B::C (b,c): If a current passes through any conductor, net charge through any section is zero. Hence, electric field in the vicinity of the conductor is zero. We can observe using Biot and Savitru law that the magnetic field at the axis of the cylinderical wire will be zero. Electric field at the axis of the wire cannot be zero otherwise the current through the wire cannot flow. |
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| 12. |
The wire shown in Fig. carries current I in the direction shown. The wire consists of a very long, straight section, a quarter- circle with radius R, and another long, straight section. What are magnitude and direction of net magnetic field at the center of curvature of quarter-circle section (point P)? |
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Answer» The contributions from the straight segments is zero since `dveclxxvecr=0` The magnetic field from the curved wire is just one quarter of a full loop: `B=1/4((mu_0I)/(2R))`, and is out of the page. |
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| 13. |
A straight section PQ of a circuit lies along the x-axis from `x=-(a)/(2)` to `x=(a)/(2)` and carries a steady current i. The magnetic field due to the section PQ at a point x =+a will beA. proportional to aB. proportional to `a^(2)`C. proportional to 1/aD. zero |
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Answer» Correct Answer - D |
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| 14. |
Two similar coils are kept mutually perpendicular such that their centres coincide. At the centre, find the ratio of the magnetic field due to one coil and the resultant magnetic field by both coils, if the same current is flownA. `1:sqrt(2)`B. `1:2`C. `2:1`D. `sqrt(3):1` |
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Answer» Correct Answer - A |
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| 15. |
The field due to a wire of n turns and radius r which carries a current I is measure on the axis of the coil at a small distance h form the centre of the coil. This is smaller than the field at the centre by the fraction:A. `3/2 (h^2)/(r^2)`B. `2/3 (h^2)/(r^2)`C. `3/2 (r^2)/(h^2)`D. `2/3 (r^2)/(h^2)` |
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Answer» Correct Answer - A (a) The magnetic field on the axis of a coil carrying current I. having n turns, radius r and at a distance h from the centre of the coil, is given by `B=(mu_0)/(4pi)xx(2piNIr^2)/((r^2+h^2)^(3//2))....(i)` The field at the centre is given by `B_C=(mu_0)/(4pi)xx(2piNI)/r....(ii)` `:. B/(B_C)=(r^3)/((r^2+h^2)^(3//2))=(r^3)/(r^3[1+(h^2)/(r^2)]^(3//2))=[1-3/2 (h^2)/(r^2)]` We have to find: `f=(B_C-B)/(B_C)=1-B/(B_C)=3/2 (h^3)/(r^2)` |
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| 16. |
A coil carrying a heavy current and having large number of turns mounted in a N-S vertical plane and current flow in clockwise direction. A small magnetic needle at its center will have its north pole inA. East -north directionB. West-north directionC. East-south directionD. West-south direction |
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Answer» Correct Answer - B |
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| 17. |
An electron is revolving round a proton, producing a magnetic field of `16 weber//m^(2)` in a circular orbit of radius `1 Å. Its angular velocity will beA. `10^(17)` rad/secB. `1//2pixx10^(12)` rad/secC. `2pi xx 10^(12)` rad/secD. `4pi xx 10^(12)` rad/sec |
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Answer» Correct Answer - A |
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| 18. |
Two circular coils X and Y, having equal number of turns and carrying currents in the same sense, subtend same solid angle at point O. If the smaller coil X is midway between O and Y and if we represent the magnetic induction due to bigger coil Y at O as `B_y` and the due to smaller coil X at O as `B_x`,then find the ratio `B_x//B_y`. A. `(B_(y))/(B_(x))=1`B. `(b_(y))/(B_(x))=2`C. `(B_(y))/(B_(x))=(1)/(2)`D. `(B_(y))/(B_(x))=(1)/(4)` |
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Answer» Correct Answer - B |
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| 19. |
A current `i` ampere flows in a circular arc of wire whose radius is `R`, which subtend an angle `3pi//2` radian at its centre. The magnetic induction `B` at the centre is A. `(mu_(0)i)/(R)`B. `(mu_(0)i)/(2R)`C. `(2mu_(0)i)/(R)`D. `(3mu_(0)i)/(8R)` |
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Answer» Correct Answer - D |
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| 20. |
A current `i` ampere flows along the inner conductor of a coaxial cable and returns along the outer conductor fo the cable, then the magnetic induction at any point outside the conductor at a distance `r` metre from the axis isA. `infty`B. zeroC. `(mu_(0))/(4pi) (2i)/r`D. `(mu_(0))/(4pi) (2pii)/(r)` |
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Answer» Correct Answer - B |
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| 21. |
Statement 1: A direct current flows through a metallic rod. It produces magnetic field only outside the rod. Statement 2: The charge carries flow through whole of the cross section.A. If both Statement 1 and Statement 2 are true, the Statement 2 is the correct explanation of Statement1.B. If both Statement 1 and Statement 2 are true, Statement 2 is not the correct explanation of Statement 1.C. If Statement 1 is true, Statement 2 is false.D. If Statement 1 is false, Statement 2 is true. |
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Answer» Correct Answer - D (d) Magnetic field exist both inside and outside. |
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| 22. |
Two long, parallel wires are separated by a distance of `0.400m` (as shown in Fig.) The current `I_1 and I_2` have the direction shown. (a) Calculate the magnitude of the force exerted by each wire on a 1.20m length of the other. Is the force attractive or repulsive ? (b) Each current is doubled, so that `I_1` becomes 10.0A and `I_2` becomes 4.00A. Now, what is the magnitude of the force that each wire exerts on a 1.20m length of the other? |
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Answer» (a) `F=(mu_0I_1I_2L)/(2pir)` `=(mu_0(5.00A)(2.00A)(1.20m))/(2pi(0.400m))=6.00xx10^-6N` The force is repulsive since the currents are in opposite directions. (d) Doubling the current make the force increase by a factor of 4 to `F=2.40xx10^-5N` |
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| 23. |
Two thin long wires carry currents `I_1 and I_2` along x- and y-axes, respectively, as shown in Fig. Consider the points only in x-y plane. A. Magnetic field is zero at last at one point in each quadrantB. Magnetic field can be zero somewhere in the first quadrantC. Magnetic field can be zero somewhere in the second quadrantD. Magnetic field in non-zero in second quadrant |
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Answer» Correct Answer - B::D (b,d) : In first and third quadrant, direction of magnetic field is opposite due to the two wires. So, magnetic field can be zero only in first and third quandrants not in second and fourth. Hence, (b) and (d) are correct. |
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| 24. |
two infinite wires, carrying currents `i_1 and i_2`., are lying along x-and y-axes, as shown in the x-y plane. Then, A. locus of points where B is zero is a circleB. locus of points where B is zero is a lineC. B decay heperbolically along any line parallel to x-axisD. B decay heperbolically along any line parallel to y-axis |
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Answer» Correct Answer - B (b) Magnetic field may be zero in first and third quadrants. `(mu_0I)/(2piy)=(mu_I_2)/(2pix)` represents a straight line passing through the origin. |
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| 25. |
Five very long, straight insulated wires are closely bound together to form a small cable. Currents carried by the wires are: `I_1=20A, I_2=-6A, I_3=12A, I_4=-7A, I_5=18A`. (Negative currents are opposite in direction to the positve.) The magnetic field induction at a distance of 10cm from the cable is (current enters at A and leaves at B and C as shown)A. `5muT`B. `15muT`C. `74muT`D. `128muT` |
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Answer» Correct Answer - C (c) Net current is `(20-6+12-7+18)A`, i.e., 37A. `r=10/100m=1/10m` `B=(mu_0I)/(2pir)=(4pixx10^-7xx37xx10)/(2pixx1)T=74xx10^-6T` `=74muT.` |
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| 26. |
Find the values of `ointvecB.vec(dl)` for the loops `L_1, L_2 and L_3` in Fig. (The sense of `vec(dl)` is mentioned in the figure) |
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Answer» For `L_1, ointvecB.dl=mu_0(I_1-I_2)`. Here `I_1` is taken postive because magnetic lines of forces produced by `I_1` is anti-lockwise as seen from top. `I_2` produces lines of `vecB` in clockwise sense as seen from top. The sense of `vecdl` is anticlockwise as seen form top. For `L_2: ointvecB.vecdl=mu_0(I_1-I_2+I_4)` For `L_3: oint vecB.vecdl=0` |
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| 27. |
An infinitely long straight wire carrying a current `I_1` is partially surrounded by a loop as shown in Fig. The loop has a length L, radius R, and carries a current `I_2` The axis of the loop coincides with the wire. Calculate the force exerted on the loop. |
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Answer» Net force on circular parts will be zero because magnetic force produced by `I_1` on these parts will be tangent at any point. On straight parts, forces will be in same direction. Net force=`2[(mu_0)/(4pi) (2I_1I_2)/R L]=(mu_0I_1I_2L)/(piR)` |
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| 28. |
Wire 1 in Fig. is oriented along the y-axis and carries a steady current `I_1`. A rectangular loop located to right of the wire and in the x-y plane carries a current `I_1`. Find the magnetic force exerted by wire (1) on the top wire (1) on the top wire of length b in the loop, labeled "wire (2)" in the figure. |
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Answer» Consider the force exerted by wire 1 on a small segment ds of wire (2). This force is given by `dvecF=IdvecsxxvecB`, where `I=I_2` `vecB` is the magnetic filed created by the current in wire (1) at the position of `dvecs`. The field at a distance x from wire (1) is `B=(mu_0I_1)/(2pix)(-hatk)` where the unit vector `-hatk` is used to idicate that the field due to the current in wire (1) at the position of `dvecs` points into the page. Because wire (2) is along the x-axis, `dvecs=dxhati`, we find that `dF_B=(mu_0I_1I_2)/(2pix)[hatixx(-hatk)]dx= (mu_0I_1I_2)/(2pi) (dx)/x hatj` Integrating over the limits `x=a` to `x=a+b` gives `F_B=(mu_0I_1I_2)/(2pi) lnx]_a^(a+b) hatj` `(mu_0I_1I_2)/(2pi) ln(1+b/a)hatj` The force on wire (2) points in the positive y-direction, as indi- cated by the unit vector `hatj` and as shown in the figure. |
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| 29. |
Two thick wires and two thin wires, all of the same materais and same length from a square in the three differenct ways, `P,Q` and `R` as shwon in figure with current connection shown, the magneitc feidl at the centre of the square is zero in cases. A. In p onlyB. In P and Q onlyC. In Q and R onlyD. P and R only |
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Answer» Correct Answer - D |
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| 30. |
Two long, straight, parallel wires are 1.00m apart (as shown in Fig). The upper wire carries a current `I_1` of 6.00A into the plane of the paper. What is the magnitude of the net field at Q?A. `2.13xx10^-6T`B. `4.26xx10^-6T`C. `1.21xx10^-6T`D. `5.30xx10^-6T` |
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Answer» Correct Answer - A (a) The field at Q point on to the right and has mangitude `B_Q=(mu_0)/(2pi)((I_1)/(r_1)-(I_2)/(r_2))=(mu_0)/(2pi)((6.00A)/(0.500m)-(2.00A)/(1.50m))` `=2.13xx10^-6T` |
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| 31. |
Two long, straight, parallel wires are 1.00m apart (as shown in Fig). The upper wire carries a current `I_1` of 6.00A into the plane of the paper. What must the magnitude of the current `I_2` be for the net field at point P to be zero?A. `3.00A`B. `sqrt3A`C. `2.00A`D. `1.00A` |
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Answer» Correct Answer - C (c) If the magnetic field at point P is zero, form the figure, the current must be out of the page, in order to cancel the field from `I_1`. Also `B_1=B_2 implies (mu_0I_1)/(2pir_1)=(mu_0I_2)/(2pir_2)` `I_2=I_1(r_2)/(r_1)= (6.00A) (0.500m)/(1.50m)=2.00A` |
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| 32. |
Find magnetic field at point P shown in Fig. point P is on the bisector of angle between the wires. |
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Answer» Assume `PO=x, so PQ=xsin(alpha//2)` Magnetic field B at P due to either segment of wire is `B=(mu_0I)/(4pix sin(alpha//2))[1+cosalpha/2]` Net magnetic field at P is `B_("net")=(mu_0I)/(2pix sin(alpha//2))[1+cos(alpha//2)]` |
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| 33. |
A pair of stationary and infintely long bent wires are placed in the ` XY` planes as shown in fig. The wires carry currents of ` I = 10 amperes` each as shown . The segments `P and Q` are parallel to the ` Y-axis` such that ` OS = OR = 0.02 m`. Find the magnitude and direction of the magnetic induction at the origin `O`. |
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Answer» As point O is along the length of segments L and M, so the field at O due to these segements will be zero. Also, point O is near one end of a long wire. The resultant field at O, `B_R=B_P+B_Q`, this field will be into the plane of paper. `implies B_R=(mu_0)/(4pi) I/(RO)+(mu_0)/(4pi) 1/(SO)` But `RO=SO=0.02m` Hence, `B_R=2xx(mu_0)/(4pi)xx10/0.02=2xx10^-7 10/0.02=10^-4 Wbm^-2` |
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| 34. |
Two identical current carrying coaxial loops, carry current I in an opposite sense. A simple amperian loop passes through both of them once. Calling the loop as C,A. `ointB.dl=mu_(0)I`B. the value of `ointB.dl=+-2mu_(0) I` is independent of sense of C.C. there mat be a point on C where, B and dI are perpendicularD. B vanishes everywhere on C. |
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Answer» Correct Answer - B::C |
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| 35. |
A long wire bent as shown in fig. carries current I. If the radius of the semicircular portion is a, the magnetic field at centre C is A. `(mu_0I)/(4a)`B. `(mu_0I)/(4pia)sqrt(pi^2+4)`C. `(mu_0I)/(4a)+(mu_0I)/(4pia)`D. `(mu_0I)/(4pia)sqrt((pi^2-4))` |
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Answer» Correct Answer - B (b) `B_1=(mu_0)/(4pi)xx(2piI)/axx1/2 ("due to semicircular part")` `B_2=(mu_0)/(4pi)xx(2I)/a ("due to parallel parts of currents")` These two fields are at right angles to each other. Hence, resultant field `B=sqrt(B_1^2+B_2^2)=(mu_0I)/(4pia)sqrt(pi^2+4)` |
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| 36. |
Three identical long solenoid P,Q and R are connected to each other as shown in fig. If the magnetic field at the centre of P is 2.0T, what would be the field (inT) at the centre of Q? Assume that the field due to any solenoid is confined within the volume of that solenoid only. |
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Answer» Correct Answer - `(1)` (1) As the solenoids are identical , the currents in Q and R will be the same and will be half the current in P. The magnetic field within a solenoid is given by `B=mu_0ni`. Hence the field in Q will be equal to the field in R and will be half the field in P, i.e., 1.0T. |
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| 37. |
A square conducting loop of side length L carries a current I.The magnetic field at the centre of the loop isA. independent of LB. proportional LC. inversely proportional to LD. linearly proportional to L |
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Answer» Correct Answer - C (c) Magnetic field at the centre due to either arm `B_1=(mu_0)/(4pi)xxI/(L//2)[sin45^@+sin45^@]` `(mu_0)/(4pi)xx(2sqrt2l)/L` Field at centre due to the four arms of the square `B=4B_1=4xx(mu_0)/(4pi)xx(2sqrt2l)/L` i.e.,`Bprop 1/L` |
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| 38. |
A current of `1//(4pi)` ampere is flowing in a long straight conductor. The line integral of magnetic induction around a closed path enclosing the current carrying conductor isA. `10^-7Wbm^-1`B. `4pi^-7Wbm^-1`C. `16pi^2xx10^-7Wbm^-1`D. zero. |
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Answer» Correct Answer - A (a) `ointvecB.dvecl=mu_0I=4pixx10^-7xx1/(4pi)=10^-7Wbm^-1` |
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| 39. |
Let `B_P and B_Q` be the magnetic field produced by wire P and Q which are placed symmetrically in a rectangular loop ABCD. Current in wire P is directed I inward and in Q is 2I outward. If `int_A^B vecB_Q.dvecl=2mu_0T-m` `int_D^B vecB_P.dvecl=-2mu_0T-m` `int_A^B vecB_P.dvecl=-mu_0T-m` Then find I. |
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Answer» `PtoAB, CD,DA,BC` `QtoAB, CD,DA,BC` Given `int_A^B vecB_Q.dvecl=2mu_0T-m` `int_D^B vecB_P.dvecl=-2mu_0T-m` `int_A^B vecB_P.dvecl=-mu_0T-m` we know `int vecB_(n et). Dvecl=mu_0(2I-I)=mu_0I` Using superpostion taking wire P only `intvecB_P.dvecl=-mu_0I and int vecB_Q.dvecl=mu_0 2I` `int_A^B vecB_Pdvecl+int_B^C vecB_Pdvecl+int_A^D vecB_Pdvecl+int_D^A vecB_Pdvecl=-mu_0I` From symmetry it is clear `int_B^C vecB_Pdvecl=int_D^A vecB_Pdvecl` As `int_A^B vecB_Qdvecl=2mu_0I` Hence, `int_B^C vecB_Pdvecl=-mu_0I` Also, `int_D^A vecB_Q dvecl=-2mu_0I` Then, `int_A^D vecB_Q dvecl=4 mu_0I=int vecB_Qdvecl` `int_A^B vecB_P dvecl=-mu_0I implies int_D^C vecB_Q dvecl=2mu_0I` `int vecB_("net") dvecl=int vecB_Pdvecl+int vecB_Q dvecl=mu_0I` `=[-mu_0I-mu_0I-2mu_0I-2mu_0I]` `+[2mu_0I+2mu_0I+4mu_0I+4mu_0I]` `=-6mu_0I+12mu_0I=mu_0(6I)` Hence, `I=6A`. |
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| 40. |
Figure shows a long wire bent at the middle to form a right angle. Show that the magnitudes of the magnetic field at point P, Q,R and S are equal and find this magnitude. |
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Answer» `|B_P|=(mu_0i)/(4pid), |B_R|=(mu_0i)/(4pid)` At P and R, the field due to vetical part is zero and the points are equidistant form the horizontal wire. Similarly, the point S and Q are equidistant from the vertical wire and field due to horizontal wire is zero. Hence `|B_Q|=|B_S|=(mu_0i)/(4pid)` `implies |B_P|=|B_Q|=|B_R|=|B_S|=(mu_0i)/(4pid)` |
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| 41. |
Four very long, current carrying wires in the same plane intersect to form a square `40.0 cm` on each side as shown in figure. Find the magnitude and direction of the current `I` so that the magnetic field at the centre of square is zero. Wires are insulated from each other. A. `4.0A` toward the bottomB. `2.0A` toward the bottomC. `2.5A` toward the bottomD. `2.6A` toward the bottom |
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Answer» Correct Answer - B (b) `BpropI`, because distance of centre for all wires is zero. `B_1=kI_1=10k ox, B_2=kI_2=8k ox` `B_3=kI_3=20k odot, B_4=kI_4` Now `B_1+B_2+B_3+B_4=0` `implies (10K+8K-20K)ox+kI_4=0` `implies kI_4=2kox` `implies I_4=2A` towards the bottom. |
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| 42. |
Find the magnitude and direction of magnetic field at point P due to the current carrying wire as shown in |
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Answer» `B=(mu_0I)/(4piR)[sin theta_1+sin theta_2]` Here, `theta_1=-30^@, theta_2=60^@`. Putting these values, we get `B=(mu_0I)/(4piR)[-1//2+sqrt3//2]=(mu_0I)/(8piR)(sqrt3-1)` |
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| 43. |
A part of a long wire carrying a current `i` is bent into a circle of radius `r` as shown in figure. The net magnetic field at the centre `O` of the circular loop is A. `(mu_(0)i)/(4r)`B. `(mu_(0)i)/(2r)`C. `(mu_(0)i)/(2pir)(pi+1)`D. `(mu_(0)i)/(2pir)(pi-1)` |
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Answer» Correct Answer - C |
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| 44. |
The magnetic field at O due to current in the infinite wire forming a loop as a shown in Fig. A. `(mu_0I)/(2pid) (cos phi_1+cos phi_2)`B. `(mu_0)/(4pi) (2I)/d (tan theta_1+tan theta_2)`C. `(mu_0)/(4pi) I/d (sin phi_1+sin phi_2)`D. `(mu_0)/(4pi) I/d (cos theta_1+cos theta_2)` |
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Answer» Correct Answer - A (a) Using `B=(mu_0i)/(4pia)[sin theta_1+sin theta_2]` But ` theta_1+phi_1=90^@ or theta_1=90^@-phi,` `sin theta_1= sin(90^@-phi_1)=cos phi_1` Similarly, `sin theta_2=cos phi_2` `B_(n et)=(mu_0I)/(2pia)(cos phi_1+cos phi_2)` |
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| 45. |
In the following figure a wire bent in the form of a regular polygon of `n` sides is inscribed in a circle of radius `a`. Net magnetic field at centre will be A. `(mu_(0)i)/(2pia)"tan"(pi)/(n)`B. `(mu_(0)ni)/(2pia)"tan"(pi)/(n)`C. `(2)/(pi)(ni)/(a)mu_(0)"tan"(pi)/(n)`D. `(ni)/(2a)mu_(0)"tan"(pi)/(n)` |
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Answer» Correct Answer - B |
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| 46. |
A staright wire of length `(pi^(2))` meter is carrying a current of `2A` and the magnetic field due to it is measured at a point distant `1 cm` from it. If the wire is to be bent into a circles and is to carry the same current as before, the ratio of the magnetic field at its centre to that obtained in the first case would beA. `50:1`B. `1:50`C. `100:1`D. `1:100` |
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Answer» Correct Answer - D |
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| 47. |
In Fig, find the magnetic field at point P. The loop is lying in x-y plane. |
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Answer» Magnetic field due to circular loop, `vecB_1=(mu_0i)/(2a) (120^@)/(360^@) (hatk)=(mu_0i)/(6a)hatk`. Due to straight wire, `vecB_2=(2mu_0i)/(4pia) xx2sin60(-hatk)=(sqrt3)/2 (mu_0i)/(pia)(-hatk)` `:. vecB_(n et)=(sqrt3)/2 (mu_0i)/(pia)(-hatk)+ (mu_0i)/(6a)(hatk)=(mu_0i)/(2pia)(sqrt3-pi/3)(-hatk)` |
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| 48. |
Two long, straight wires, one above the other, are separated by a distance 2a and are parallel to the x-axis. Let the y-axis be in the plane of the wires in the direction form the lower wire to the upper wire. Each wire carries current I in the +x-direction. What are the magnitude and direction of the net magnetic field of the two wires at a point in the plane of wires: (a) midway between them? (b) at a distance a above the upper wire? (c) at a distance a below the lower wire? |
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Answer» (a) At the point exactly midway between the wires, the two magnetic fields are in opposite directions and cancel. (b) At a distance a above the top wire, the magnetic fields are in the same direction and add up `vecB=(mu_0I)/(2pir_1)hatk+(mu_0I)/(2pir_2)hatk=(mu_0I)/(2pia)hatk+(mu_0I)/(2pi(3a))hatk=(2mu_0I)/(3pia)hatk` (c) At a distance a below the lower wire, magnitude magnetic field is same as in part (b) but is in the opposite direction `vecB=-(2mu_0I)/(3pia) hatk` |
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| 49. |
Two semicircles shown in Fig. have radii a and b. Calculate the net magnetic field (magnitude and direction) that the current in the wires produces at point P. |
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Answer» `B=(B_a-B_b)=1/2((mu_0I)/2) (1/a-1/b)` `B=(mu_0I)/(4a)(1-a/b)`, out of the page. |
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| 50. |
Find the magnetic force on loop PQRS due to the wire. |
| Answer» `F_(res) =(mu_0I_1I_2)/(2pia)a(-hati)+(mu_0I_1I_2)/(2pi(2a))a(hati)=(mu_0I_1I_2)/(4pi)(-hati)` | |