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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.
| 101. |
A particle of mass `m`and charge `q`, moving with velocity `v` enters Region `II` normal to the boundary as shown in the figure. Region `II` has a uniform magnetic field `B` perpendicular to the plane of the paper . The length of the region `II` is `l`. Choose the correct choice(s).A. The particle enters regions III only if its velocity `V gt (qlB)/(m)` .B. The particle enters regions III only if its velocity `V gt (qlB)/(m)` .C. Path length of the particle in region II is maximum when velocity `V = (qlB)/(m)`D. Time spent in region II is same for any velocity `V` as long as the particle returns to region I . |
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Answer» Correct Answer - `A,C,D` In region II the particle follows a circular path of radius ` r = (mv)/(qB)` Therefore, the particle can enter III,ifr gt l i.e if `v gt (qBl)/(m)` In regin II the maximum path length is `r =l` which gives `v =(qBl)/(m)` The time period of the circular motion is ` T = (2pir)/(v) = (2pi)/(v) xx (mv)/(qB) = (2pi)/(qB)` The particle will return to region I if the time spent by in region II is `(T)/(2) =(pim)/(qB)` which is indepedndent of the velocity Hence the correct choices are `(A)` `(C)` and `(D)` . |
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| 102. |
A galvanometer can be converted into a voltmeter of certain range by connecting a resistance of `880Omega` in series with it. When the resistance of `420Omega` is connected in series, the range becomes half. Find the resistance of galvanometer. |
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Answer» Correct Answer - `40Omega` Let G be the resistance of galvanometer. If a resistance R is connected in series of galvanometer it works as a voltmeter of range 0 to V volt. The current for full scale deflection of a voltmeter is `I_g=(V)/(R+G)` In first case, `I_g=(V)/(880+G)` In second case, `I_g=(V//2)/(420+G)=(V)/(2(420+G))` `:. (V)/(880+G)=(V)/(2(420+G))` or `880+G=840+2G` or `G=40Omega`. |
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| 103. |
A galvanometer reads `5*0V` at full scale deflection and is graded according to its resistance per volt at full scale deflection as `5000OmegaV^-1`. (i) How will you convert it into a voltmeter that reads `20V` at full scale deflection? (ii) Will it still be graded `5000OmegaV^-1`? (iii) Will you prefer this voltmeter to one that is graded `2000OmegaV^-1`? |
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Answer» Here, resistance per volt=`5000OmegaV^-1`. It means for `1` volt pot. difference, the resistance of galvanometer is `5000Omega`. So current for full scale deflection, `I_g=1/5000A=0*2xx10^-3A` (i) In order to convert it into a voltmeter and range 0 to `20V`, let a resistance R be connected in series with it. Then on applying an extra potential difference `=20V-5V=15V`, the current potential it for full scale deflection is again `0*2xx10^-3A` i.e., `I_g=0*2xx10^-3A`. `:. RI_g=15` or `R=15/I_g=(15)/(0*2xx10^-3)=75000Omega` Thus, to convert a given voltmeter (of range 0 to 5V) into a voltmeter (of range 0 to 20V) a resistance of `75000Omega` should be connected in series. With the given meter. (ii) Original resistance of voltmeter `=5000OmegaV^-1xx5V=25000Omega` Total resistance after conversion `=25000+75000=100,000Omega` Resistance per volt of new meter `=(100,000)/(20)=5000OmegaV^-1` i.e. it has the same grading as before. (iii) If greater is the resistance per volt of a meter, the lesser will be the current drawn from the circuit by it. Due to it, it works better. That is why this meter (graded `5000OmegaV^-1`) is more accurate that the one graded as `2000OmegaV^-1`. |
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| 104. |
A resistance of `1980Omega` is connected in series with a voltmeter, after which the scale division becomes 100 times larger. Find the resistance of voltmeter. |
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Answer» Let R be the resistance of voltmeter. Let n be the number of divisions in the voltmeter. The voltage `(V)` recorded by each division of voltmeter when current `i_g` flows through it is `i_gR//n=V` …(i) When resistance is connected in series with voltmeter, then `i_g(R+1980)//n=100V` ...(ii) Dividing (ii) by (i), we get `R+1980=100R` or `R=1980//99` `=20Omega` |
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| 105. |
Wires `1 and 2` carrying currents `i_(1) and i_(2)` respectively are inclined at an angle `theta ` to each other. What is the force on a small element `dl` of wire `2` at a distance of `r` from wire 1 ( as shown in figure) due to the magnetic field of wire 1`? A. `(mu_(0))/(2pir) i_(1) i_(2) dl tan theta`B. `(mu_(0))/(2pir) i_(1) i_(2) dl sin theta`C. `(mu_(0))/(2pir) i_(1) i_(2) dl cos theta`D. `(mu_(0))/(4pir) i_(1) i_(2) dl sin theta` |
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Answer» Correct Answer - C |
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| 106. |
Two parallel beams of protons and electrons, carrying equal currents are fixed at a separation d. The protons and electrons move in opposite directions. There is a point P on the straight perpendicular line joining the two beams at a distance x from one beam. The magnetic field at this point is B. If B is plotted against x, it can be represented by the curve.A. B. C. D. |
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Answer» Correct Answer - C |
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| 107. |
A and B are two conductors carrying a current i in the same direction x and y are two electron beams moving in the same direction. Then A. There will be repulsion between `A` and `B` attraction between `x` and `y`B. There will be attraction between `A` and `B`, repulsion between `x` and `y`C. There will be repulsion between `A` and `B` and also `x` and `y`D. There will be attraction between `A` and `B` and also `x` and `y` |
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Answer» Correct Answer - B |
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| 108. |
Two parallel beams of electrons moving in the same direction produce a mutual forceA. Of attraction in plane of paperB. Of repulsion in plane of paperC. Upwards perpendicular to plane of paperD. Downwards perpendicular to plane of paper |
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Answer» Correct Answer - B |
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| 109. |
Net magnetic field at the centre of the circle `O` due to a currying loop as shown in figure is `(thetalt180^(@))` .A. zeroB. perpendicular to paper inwardsC. perpendicular to paper outwardsD. is perpendicular to paper inwards if `theta le 90^(@)` and perpendicular to paper outward of `90^(@)letheta lt180^(@)` . |
| Answer» Correct Answer - C | |
| 110. |
The unit of electric current "ampere" is the current which when flowing through each of two parallel wires spaced `1 m` apart in vacuum and of infinite length will give rise to a force between them equal toA. `1N//m`B. `2xx10^(-7)N//m`C. `1xx10^(-2)N//m`D. `4pi xx10^(-7)N//m` |
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Answer» Correct Answer - B |
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| 111. |
In the frame work of wires shown in figure, a current I ampere is flowing. Find the magnetic field at O. |
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Answer» Let `hatn` be the unit vector perpendicular to the plane of frame work upwards. Total magnetic field at P is `vecB_0=vecB_1+vecB_2+vecB_3+vecB_4` `=(mu_0)/(4pi)I/r_1(2pi-alpha)(-hatn)+0+(mu_0)/(4pi)I/r_2alpha(-hatn)+0` `=(mu_0)/(4pi)I((2pi-alpha)/(r_1)+alpha/r_2)(-hatn)` `B_0=(mu_0)/(4pi)I((2pi-alpha)/(r_1)+alpha/r_2)` The direction of `vec(B_(0))` is perpendicular to the plane of framework of wire directed inwards. |
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| 112. |
A solenoid of 1.5 metre length and 4.0 cm diameter posses 10 turn per cm. A current of 5 ampere is flowing through it. The magnetic induction at axis inside the solenoid isA. `2pixx10^(-3)` TeslaB. `2pixx10^(-5)` TeslaC. `4pixx10^(-2)` GaussD. `2pixx10^(-5)` Gauss |
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Answer» Correct Answer - A |
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| 113. |
For a positively charged particle moving in a `x-y` plane initially along the `x-axis` , there is a sudden change in its path due to the presence of electric and//or magnetic fields beyond `p` . The curved path is shown in the ` x- y `plane and is found to be non - circular. Which one of the following combinations is possible ? A. `vecE=0,vecB=bhati+chatk`B. `vecE=ai,vecB=chatk+ahati`C. `vecE=0,vecB=chatj+bhatk`D. `vecE=ai,vecB=chatk+bhatj` |
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Answer» Correct Answer - B |
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| 114. |
An ionized gas contains both positive and negative ions . If it is subjected simultaneously to an electric field along the ` +x` - direction and a magnetic field along the ` +y` - direction and the negative ions towardws `-y` - directionA. Positive ions deflect towards `+y` direction and negative ions towards `–y` directionB. All ions deflect towards` +y` directionC. All ions deflect towards `–y` directionD. Positive ions deflect towards `–y` direction and negative ions towards `+y `direction |
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Answer» Correct Answer - C |
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| 115. |
A metallic block carrying current `I` is subjected to a uniform magnetic induction `vec(B) as shown in Figure . The moving charges experience a force ` vec(F) given by ….. Which results in the lowering of the potential of the face ……. Assume the speed of the carries to be `v` . A. `eVbhatk, ABCD`B. `eVBhatk, EFGH`C. `-eVBhatk,ABCD`D. `-eVBhatk, EFGH` |
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Answer» Correct Answer - A |
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| 116. |
An electric current of I ampere is flowing in a long conductor CG as shown in figure. Find the magnitude and direction of magnetic induction at the centre O of circular part. |
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Answer» Correct Answer - `(mu_0I)/(2r)(1/pi+1)` acting `_|_r` to loop upwards `vecB_0=vecB_(CG)+vecB_(DEF)` `=(mu_0)/(4pi)(2I)/(r)hatn+(mu_0)/(4pi)(2piI)/(r)hatn=(mu_0I)/(2r)[1/pi+1]hatn` where `hatn` is a unit vector perpendicular to loop upwards. |
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| 117. |
For a positively charged particle moving in a `x-y` plane initially along the `x-axis` , there is a sudden change in its path due to the presence of electric and//or magnetic fields beyond `p` . The curved path is shown in the ` x- y `plane and is found to be non - circular. Which one of the following combinations is possible ? A. `oversetrarrE = 0 , oversetrarrB = bhatj + c hatk`B. `oversetrarrE = ahati , oversetrarrB = chatk + ahati`C. `oversetrarrE = 0 , oversetrarrB = cchatj + bhatk`D. `oversetrarrE = ahati , oversetrarrB = chatk + bhatj` |
| Answer» Correct Answer - `B` | |
| 118. |
A charge moving with velocity `v` in `X`-direction is subjected to a field of magnetic induction in the negative `X`-direction. As a result, the charge willA. Remain unaffectedB. Start moving in a circular path `Y-Z` planeC. Retard along `X`-axisD. Move along a helical path around `X`- axis |
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Answer» Correct Answer - A |
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| 119. |
A chamber is maintained a uniform magnetic field of `5xx10^-3T`. An electron with a speed of `5xx10^7ms^-1` enters the chamber in a direction normal to the field. Calculate (i) radius of the path (ii) frequency of revolution of the electron. Charge of electron `=1*6xx10^(-19)C`, Mass of electron `=9*1xx10^(-31)kg` |
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Answer» Correct Answer - (i) `5*7cm` (ii) `1*4xx10^8Hz` Here, `B=5xx10^-3T`, `v=5xx10^7ms^-1`, `e=1*6xx10^(-19)C`, `m=9*1xx10^(-31)kg` (i) Radius of the circular path, `r=(mv)/(Be)=((9*1xx10^(-31))xx(5xx10^7))/((5xx10^-3)xx(1*6xx10^(-19)))` `=5*7xx10^-2m` `=5*7cm` (ii) Frequency of revolution, `v=(eB)/(2pim)=((1*6xx10^(-19))xx(5xx10^-3))/(2xx3*14xx(9*1xx10^(-31))` `=1*4xx10^8Hz` |
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| 120. |
Calculate the magnetic field due to a circular coil of `250` turns and of diameter `0*1m`, carrying a current of `7A` (i) at the center of the coil (ii) at a point on the axis of the coil at a distance `0*12m` from the center of the coil. |
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Answer» Correct Answer - (i) `2*2xx10^-2T` (ii) `1*25xx10^-3T` (i) `B=(mu_0)/(4pi)(2pinI)/(r)` `=(10^-7xx2xx(22//7)xx250xx7)/((0*1//2))=2*2xx10^-2T` (ii) `B=(mu_0)/(4pi)(2pinIa^2)/((a^2+x^2)^(3//2))` `=(10^-7xx2xx(22//7)xx250xx7xx(0*1//2)^2)/([(0*1//2)^2+(0*12)^2]^(3//2))` `=1*25xx10^-3T` |
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| 121. |
An electron enters electric field of `10^4V//m` perpendicular to the field with a velocity of `10^7m//s`. Find the vertical displacement of electron after 2 milliseconds. Mass of electron `=9*1xx10^(-31)kg`, charge on electron `=1*6xx10^(-19)C`. |
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Answer» Correct Answer - `3*51xx10^9m` `y=1/2at^2=1/2(Ee)/(m)t^2` `=(10^4xx(1*6xx10^(-19))xx(2xx10^-3)^2)/(2xx9*1xx10^(-31))` `=3*51xx10^9m` |
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| 122. |
The dipole moment of a coil of 200 turns carrying a current of `3A` is `9*24Am^2`. What is the diameter of the coil? |
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Answer» Correct Answer - `14cm` Here, `N=200`, `I=3A`, `M=9*24Am^2d=?` If r is radius of coil, then `M=NIA=Nipi(d^2)/(4)` `d=sqrt((4M)/(piNI))=sqrt((4xx9*24)/(3*14xx200xx3))` `=0*14m=14cm` |
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| 123. |
A small coil of `N` turns has area `A` and a current `I` flows through it. The magnetic dipole moment of this coil will beA. `NI//A`B. `NI^(2)A`C. `N^(2)AI`D. `NlA` |
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Answer» Correct Answer - D |
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| 124. |
A long solenoid of length `L` has a mean diameter `D`. It has `n` layers of windings of `N` turns each. If it carries a current `‘i’` the magnetic field at its centre will beA. Proportional to `D`B. Inversely proportional to `D`C. Independent of `D`D. Proportinal to `L` |
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Answer» Correct Answer - C |
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| 125. |
Field at the centre of a circular coil of radius `r`, through which a current `I` flows isA. Directly proportional to `r`B. Inversely proportional to `l`C. Directly proportional to `l`D. Directly proportional to `I^(2)` |
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Answer» Correct Answer - C |
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| 126. |
A straight wire carrying a current `10 A` is bent into a semicircular arc of radius `5 cm`. The magnitude of magnetic field at the center isA. `1.5xx10^(-5)T`B. `3.14xx10^(-5)T`C. `6.28xx10^(-5)T`D. `19.6xx10^(-5)T` |
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Answer» Correct Answer - C |
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| 127. |
In the figure shown, the magnetic induction at the centre of the arc due to the current in portion `AB` will be A. `(mu_(0)i)/r`B. `(mu_(0)i)/(2r)`C. `(mu_(0)i)/(4r)`D. Zero |
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Answer» Correct Answer - D |
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| 128. |
A current i flows in a circular coil of radius `r`. If the coil is placed in a uniform magnetic field `B` with its plane parallel to the field, magnitude of the torque that acts on the coil isA. ZeroB. `2pi r i B`C. `pi r^(2) i B`D. `2 pi r^(2) B` |
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Answer» Correct Answer - C |
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| 129. |
If current through a circular coil flows in clockwise direction, then the direction of magnetic field at the centre of the circular coil is………………to the plane of the coil, directed……………. . |
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Answer» Correct Answer - perpendicular ; inwards, |
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| 130. |
Figure shows a rectangular current-carrying loop placed `2cm` away from a long, straight, current carrying conductor. What is the direction and magnitude of the net force acting on the loop? |
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Answer» Here, `I_1=15A, I_2=25A`, `r_1=2xx10^-2m, r_2=(2+10)xx10^-2m` Force on BC, `F_1=(mu_0)/(4pi)(2I_1I_2)/(r_1)xxl enght BC` `=10^-7xx(2xx15xx25)/((2xx10^-2))xx(25xx10^-2)` `=9*375xx10^-4N` (repulsive, away from XY) Force on DA, `F_2=(mu_0)/(4pi)(2I_1I_2)/(r_2)xx l eng th DA` `=10^-7xx(2xx15xx25)/((2+10)xx10^-2)xx25xx10^-2` `=1*5625xx10^-4N` (attractive towards XY) Net force on the loop `F=F_1-F_2` `=(9*375-1*5625)xx10^-4` `=7*8175xx10^-4N` |
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| 131. |
A circular coil of `300` turns and diameter `14cm` carries a current of `15A`. What is the magnitude of magnetic moment linked with the loop? |
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Answer» Here, `N=300`, `r=14/2cm=7xx10^-2m, I=15A` `M=NIA=NI(pir^2)` `=300xx15xx3*14(7xx10^-2)^2=69*2Am^2` |
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| 132. |
A conducting circular loop of radius r carries a constant current i. It is placed in a uniform magnetic field B such that B is perpendicular to the plane of loop. What is the magnetic force acting on the loop? |
| Answer» Correct Answer - Zero | |
| 133. |
Figure, shows an equilateral triangular loop CDE, carrying current I. Length of each side of triangle is l. If a uniform magnetic field exists parallel to side DE of loop, then find the foces acting on the three wires CD, DE and EC separately. |
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Answer» Correct Answer - `IlBsqrt3//2`, normally outwards, zero `IlBsqrt3//2`, normally inwards For current in wire CD, the angle between `vecl` and `vecB` is `120^@`, i.e., `theta=120^@`. Therefore, force on wire CD is `vecF_(CD)=I[(vec(CD))xxvecB]=IlBsin 120^@hatn` or `F_(CD)=IlBsin120^@=IlBxxsqrt3//2` Its direction according to Right Hand Rule is normally out of the plane of paper. For arm DE, the anggle between `vecl` and `vecB` is zero i.e., `theta=0^@`. Therefore, force on wire DE is `IlBsin0^@=0` For arm EC, the angle between `vecl` and `vecB` is `120^@` i.e., `theta=120^@`. Therefore, force on wire EC is `F_(CE)I(CE)Bsin120^@=IlBxxsqrt3//2` Its direction is normally into the plane of paper. |
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| 134. |
A circular loop of radius 10 cm carries a current of 15A. At its center is placed a small loop of radius 1cm with 50 turns and a current of 1A. A) what is the B-vector of the magnetic field produced by the large loop at its center? (b) What is the H-vector of the field? (c) What torque acts on the small loop when its plane is perpendicular to the plane of the large loop? |
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Answer» Correct Answer - (a) `9.4 xx 10^(-5)`T, (b) 74.96 `Am69-1)`, © `1.5 xx 10^(-6)`Nm |
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| 135. |
A conducting circular loop of radiius `r` carries a constant current `i`. It is placed in a uniform magnetic field `vec(B)_(0)` such that `vec(B)_(0)` is perpendicular to the plane of the loop . The magnetic force acting on the loop isA. `irvecB`B. `2pir ivecB`C. ZeroD. `pi ri vecB` |
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Answer» Correct Answer - C |
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| 136. |
A current carrying circular loop lies on a smooth horizontal plane. Can a uniform magnetic field be set up in such a manner that the loop turns around itself (i.e., turns about the vertical axis). |
| Answer» No, because the loop can turn around itself if the torque `tau` is acting in the vertical direction. Since, torque, `vectau=(vecMxxvecB)=I(vecAxxvecB)` and `vecM` or `vecA` of the horizontal loop is in the vertical direction, so the direction of `vectau` will be in the plane of the loop for any direction of `vecB`. | |
| 137. |
A triangular loop of side `l` carries a current `I`. It is placed in a magnetic field `B` such that the plane of the loop is in the direction of `B`. The torque on the loop isA. ZeroB. `lBl`C. `(sqrt(3))/2 Il^(2)B^(2)`D. `(sqrt(3))/4 I Bl^(2)` |
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Answer» Correct Answer - D |
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| 138. |
Assertion: A circular loop carrying current lies in `XY` plane with its centre at origin having a magnetic flux in negative `Z`-axis. Reason: Magnetic flux direction is independent of the direction of current in the conductor.A. If both assertion and reason are true and the reason is the correctexplanation of the assertion.B. If both assertion and reason are true but reason is not the correctexplanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - C |
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| 139. |
A square loop of mass m and side length a lies in xy plane with its centre at origin. It carries a current I. The loop is free to rotate about x axis. A magnetic field `vec(B)=B_(0 )= B_(0)vec(j)` is switched on in the region. Calculate the angular speed acquired by the loop when it has rotated through 90°. Assume no other force on the loop apart from the magnetic force. |
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Answer» Correct Answer - `omega=sqrt((12IB_(0))/m` |
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| 140. |
What is Meissner effect? |
| Answer» It is the phenomenon of perfect diamagnetism of superconductors. | |
| 141. |
What is magnetic susceptibility of super conductors. |
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Answer» As `mu_r=1+chi_m=0`, for superconductors, `:. chi_m=-1` |
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| 142. |
What is relative magnetic permeability of superconductors? |
| Answer» For superconductors, `mu_r=0`. | |
| 143. |
Figure shows a cube made from twelve uniform wires. Find the magnetic field at the centre of the cube, if a battery is connected between the points A and H. |
| Answer» We have the following current carrying conducting pairs AB and EH, AF and CH, AD and GH, BG and DE, GF and CD, FE and BC, for which each pair produce equal and opposite magnetic fields at the centre. As a result of it, the resultant magnetic field at the centre will be zero. | |
| 144. |
Two identical circular oils of radius `0*2m` each having 30 turns are mounted coaxially `0*2m` apart. What is the magnetic field at the centre of each coil when a current of `0*6A` is passed through both the coils (a) in the same direction (b) in the opposite directions? |
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Answer» Here, `a=0*2m`, `n=30`, `x=0*2m`, `I=0*6A` Magnetic field at the centre of each coil due to its own current is `B_1=(mu_0nI)/(2a)=((4pixx10^-7)xx30xx0*6)/(2xx0*2)` `=5*65xx10^-5T` Magnetic field at the centre of one coil due to the current in the other coil is `B_2=(mu_0)/(4pi)(2pinIa^2)/((a^2+x^2)^(3//2)` `=(10^-7xx2pixx30xx0*6xx(0*2)^2)/([(0*2)^2+(0*2)^2]^(3//2))` `=2*05xx10^-5T` (a) When the currents are in the same direction the resultant magnetic field at the centre of each coil is `B=B_1+B_2=5*65xx10^-5+2*05xx10^-5` `=7*70xx10^-5T` (b) When the currents are in the opposite directions, the resultant magnetic field at the centre of each coil is `B=B_1-B_2=5*65xx10^-5-2*05xx10^-5` `=3*60xx10^-5T` |
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| 145. |
A cathode ray tube contains a pair of parallel metal plates `1*0cm` apart and `3*0cm` long. A narrow horizontal beam of electrons with velocity of `3xx10^7ms^-1` is passed down the tube mid way between the two plates. When the potential difference of `550V` is maintained across the plates, it is found that the electron beam is so deflected that it just strikes the end of one of the plates. Find the specific charge of an electron. |
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Answer» Here, `y=1/2cm=0*5xx10^-2m`, `l=3cm=3xx10^-2m, v=3xx10^7ms^-1`, `d=1*0cm=10^-2m, V=550V` Time to cross the tube, `t=l/v=(3xx10^-2)/(3xx10^7)=10^-9s` `y=1/2at^2=1/2((Ee)/(m))t^2=1/2((V//d)e)/(m)t^2` or `e/m=(2yd)/(Vt^2)=(2xx(0*5xx10^-2)xx10^-2)/(550xx(10^-9)^2)` `=1*8xx10^(11)Ckg^-1` |
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| 146. |
How is the magnetic field inside a given solenoid made strong? |
| Answer» It is done by inserting laminated iron core inside the solenoid. | |
| 147. |
Magnetic field lines can be entirely confined within the core of a toroid, but not within a straight solenoid. Why? |
| Answer» It is so because the magnetic field induction outside the toroid is zero. | |
| 148. |
Assertion : Magnetic field interacts with a moving charge and not with a stationary charge. Reason : A moving charge produces a magnetic field.A. If both assertion and reason are true and the reason is the correctexplanation of the assertion.B. If both assertion and reason are true but reason is not the correctexplanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - A |
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| 149. |
At a specific instant emission of radioactive compound is deflected in a magnetic field . The compound can emit (i) electron (ii)protons(iii)`He^(2+ )`(iv) neutrons The emission at instant can beA. i,ii,iiiB. i,ii,iii,ivC. ivD. ii,iii |
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Answer» Correct Answer - A |
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| 150. |
A charged particle can be deflected by electric and magnetic filed both The electric force `oversetrarrF_(e) =qoversetrarrE` and magnetic force `oversetrarr(F_(m)) = oversetrarr(qv)xxoversetrarr(B) q` = charge in coulomb, oversetrarrV is velocity in m/s `oversetrarrE` is electric filed strengh in `N//C` and `B` is magnetic field strength in tesla `(T)` If a moving charged particle enters simultaneous electric and magnetic field, the net force on the charged particle is `oversetrarr(F) =q(oversetrarrE +oversetrarr(v) xx oversetrarrB)` If `oversetrarr(F) =0` the charged particle will move undeflected in simultaneous fields If electric field is along `Y` axis and charged particle moves along `X` axis the magnetic field must be .A. `(E)/(v)` along negative z-axisB. `(E)/(v)` along positive z-axisC. `(E)/(v)` along negative z-axisD. `vE` along positive z-axis . |
| Answer» Correct Answer - `B` | |