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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 solenoid of 1000 turns is wound uniformly on a glass tube `50cm` long and `10cm` diameter. The strength of magentic field at the centre of solenoid when a current of `0.1A`. Flows through it isA. `100 A//m`B. `200A//m`C. `400A//m`D. `50 A//m` |
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Answer» Correct Answer - B `H=ni` |
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| 102. |
A solenoid of length `50cm`, having `100` turns carries a current of `2*5A`. Find the magnetic field, (a) in the interior of the solenoid, (b) at one end of the solenoid. Given `mu_0=4pixx10^-7WbA^-1m^-1`.A. `3.14 xx 10^(-4)T`B. `6.28 xx 10^(-4)T`C. `1.57 xx 10^(-4)T`D. `9.42 xx 10^(-4)T` |
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Answer» Correct Answer - A Here, I = 2.5 A, l = 50 cm = 0.50 m and `n=(100)/(0.50)=200m^(-1)` `therefore" "B=(mu_(0)nI)/(2)=(4pi xx 10^(-7)xx200xx2.5)/(2)=3.14xx10^(-4)T` |
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| 103. |
A current carrying loop is placed in a uniform magnetic field. The torque acting on it does not depend uponA. area of loopB. value of currentC. magnetic fieldD. None of these |
| Answer» Correct Answer - D | |
| 104. |
If the angular momentum of an electron revolving in a circular orbit is `L`, then its magnetic moment isA. `eLm`B. `eL//m`C. `eL//2m`D. zero |
| Answer» Correct Answer - 3 | |
| 105. |
A wire length `6.28m` is bent into a circular coil of 2 turns. If a current of `0.5A` exists in the coil, the magnetic moment of the coil is, in `Am^(2) :`A. `(pi)/(4)`B. `(1)/(4)`C. `pi`D. `4pi` |
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Answer» Correct Answer - 1 `M=NIpir^(2),l=(2pir)N` |
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| 106. |
In the Bohr model of the hydrogen atom, the electron circuulates around the nucleus in a path of radius `5xx10^-11m` at a frequency of `6.8xx10^15Hz`. a. What value of magnetic field is set up at the centre of the orbit? b. What is the equivalent magnetic dipole moment?A. `12.27T`B. `10.8T`C. `13.2T`D. `13.6T` |
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Answer» Correct Answer - 4 `B=(mu_(0)n e)/(2r)` where n is the frequency |
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| 107. |
A particle of the charged `q` and `mass m` moves in a circular orbit of radius `r ` with angular speed ` omega` . The ratio of the magnitude of its magnetic moment to that of its angular momentum depends onA. ` omega and q `B. ` omega , q and m `C. ` q and m `D. ` omega and m ` |
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Answer» Correct Answer - C The angular momentum `L` of the particle is given by ` L = mr^(2)omega` where `omega = 2 pin`. :. Frequency ` n = (omega)/( 2pi)` , Further ` I = qxxn = ( omega q)/( 2 pi)` magnetic moment , `M = I A = ( omega q)/( 2 pi)xx pir^(2)` :. ` M = ( omegaqr^(2))/(2) ` S0 , `(M)/(L) = ( omega q r^(2))/ ( 2mr^(2)omega) = (q)/(2 m) ` |
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| 108. |
A charged particle of mass m enters in a uniform magnetic field B with speed V as shown. |
| Answer» Correct Answer - A(r), B(r), C(q), D(q) | |
| 109. |
A proton of mass `1.67xx10^(-27)kg` and charge `1.6xx10^(-19)c` is projected with a speed of `2xx10^(6)m//s` at an angle of `60^(@)` to the x-axis. If a uniform magnetic field of `0.104` Tesla is applied along Y-axis, the path of proton isA. a circle of radius `~=0.1m` and time perio `pi xx 10^(-7)s`B. a circle of radius `~=0.2m` and time period `2pixx10^(-7)s`C. a helix of radius `~=0.1m` and time period `2pixx10^(-7)s`D. a helix of radius `~=0.2m` and time period `4pixx10^(-7)s` |
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Answer» Correct Answer - C `v_(_|_)=v cos 60`, find `r=(mv_(_|_))/(qB),r=0.1m` Hence `t=(2pim)/(qB)=2pixx10^(-7)s`. Hence answer is `(c )` and path is helix. |
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| 110. |
An electron of charge `e` moves in a circular orbit of radius `r` round a nucleus the magnetic field due to orbit motion off the electron at the site of the nucleus is `B`. The angular velocity `omega` of the electron isA. `omega=(mu_(0)eB)/(4pir)`B. `omega=(mu_(0)eB)/(pir)`C. `omega=(4pirB)/(mu_(0)e)`D. `omega=(2pirB)/(mu_(0)e)` |
| Answer» Correct Answer - 3 | |
| 111. |
If an electron is revolving in a circular orbit of radius `0.5A^(@)` with a velocity of `2.2xx10^(6)m//s`. The magnetic dipole moment of the revolving electron isA. `8.8xx10^(-24)Am`B. `8.8xx10^(-23)Am`C. `8.8xx10^(-22)Am`D. `8.8xx10^(21)Am` |
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Answer» Correct Answer - 1 `M= i nA ` and `i=(q)/(T)` |
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| 112. |
An electron of charge `"e"` has a time period of revolution of `"T"` in a Bohr orbit of radius `"r"`. The dipole moment of the electron isA. `pi r^(2)eT`B. `pi r^(2)e//T`C. `pi r^(2)T//e`D. `T//pi r^(2)e` |
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Answer» Correct Answer - 2 `M=iA=(e)/(T)A=(e)/(T)pir^(2)` |
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| 113. |
A charged particle enters into a uniform magnetic field the parameter that remains constant isA. velocityB. momentumC. kinetic energyD. angular velocity |
| Answer» Correct Answer - 3 | |
| 114. |
A uniform electric field and a uniform magneitc field exist in a region in the same direction An electron is projected with velocity pointed in the same direction the electron willA. turn to its rightB. turn to its leftC. keep moving in the same direction but its speed will increaseD. keep moving in the same direction but its speed will decrease |
| Answer» Correct Answer - 4 | |
| 115. |
A proton and an electron are projected into a region of uniform magnetic field in a direction perpendicular to the field. If they have the same initial velocities thenA. They move in circular paths of same radiiB. They experience equal forceC. The trajectory of the electron is more curvedD. Their velocity remain equal to each other through out the motion |
| Answer» Correct Answer - C | |
| 116. |
An electron of charge`e`, revolves round in an orbit of radius `r` with a uniform angular velocity `omega`. The magnetic dipole moment of the electron in the orbit isA. `eomegar//2`B. `e omegar^(2)//2`C. `eomega^(2)r//2`D. `eomega^(2)r^(2)//2` |
| Answer» Correct Answer - B | |
| 117. |
An electron is projected parallel to electric and uniform magnetic fields acting simultaneously in the same direction. The electron.A. gains kinetic energyB. loses kinetic energyC. moves along circular pathD. moves along a parabolic path |
| Answer» Correct Answer - 2 | |
| 118. |
A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected along the direction of the fields with a certain velocity thenA. its velocity will increaseB. its velocity will decreaseC. it will turn towards left of direction of motionD. it will turn towards right of direction of motion |
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Answer» Correct Answer - B Due to electric field , it experiences force and decelerates i.e. its velocity decreases . |
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| 119. |
If a charged particle is projected perpendicular to a uniform magnetic field, then `a)` it revolves in circular path `b)` its `K.E.` remains constant `c)` its momentum remains constant `d)` its path is psiralA. only`a,b` are correctB. only `a,c,` are correctC. only`b,d` are correctD. only `a,d` are correct |
| Answer» Correct Answer - 1 | |
| 120. |
Magnetic force acting on the charged particle projected perpendicular to magnetic field is proportional to `r^(n)`, where r is radius of circular path. Find n |
| Answer» Correct Answer - 1 | |
| 121. |
If a galvanometer is shunted then among the following which statement is not trueA. effective range increases.B. equivalent resistance decreases.C. galvanometer becomes more sensitiveD. galvanometer becomes more protective. |
| Answer» Correct Answer - 3 | |
| 122. |
A maximum current point `0.5mA` can be passed through a galvanometer of resistance `20Omega` . The resistance to be connected in series to convert it in the voltmeter of range `0-5V` isA. `980Omega`B. `9980Omega`C. `990Omega`D. `9990Omega` |
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Answer» Correct Answer - 2 `R=((V)/(I_(g))-G)` |
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| 123. |
A galvanometer of resistance `20Omega` is to be shunted so that only `1%` of the current passes through it. Shunt connected is `99//20Omega`A. `9//20Omega`B. `20//99Omega`C. `2//99Omega`D. |
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Answer» Correct Answer - 3 `S=(G)/((i)/(i_(g))-1)` |
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| 124. |
A galvanometer of resistance `20Omega` is shunted by a `2 Omega` resistor. What part of the main current flows through the galvanometer ?A. `1//10` partB. `1//11 ` partC. `1//12 ` partD. `1//13` part |
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Answer» Correct Answer - 2 `i_(g)=(iS)/(G+S)` |
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| 125. |
A microammeter has as resistance of `100 Omega` and full scale range of `50 muA`. It can be used a voltmeter or as ahigher range ammeter provided a resistance is added to it. Pick the correct range and resistance combinations 50 V range with `10 kOmega` resistance in series b.`10 V` range with `200 kOmega` resistance in series c. 5mA rangw with `1Omega` resistance in parallel `10 mA` range with `1Omega` resistance in parallelA. 50 Volt range with `10 K Omega` resistance in seriesB. 5 Volt range with `200 K Omega` resistance in seriesC. `5 Ma` range with` 1 Omega` resistance in parallelD. `10 Ma` range with `1Omega` resistance in parallel |
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Answer» Correct Answer - 3 `S=(G)/((i)/(ig)-1)` |
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| 126. |
What is the correct value of Bohr magneton?A. `8.89 xx 10^(-24)A m^(2)`B. `9.27 xx 10^(-24)A m^(2)`C. `5.56 xx 10^(-24)A m^(2)`D. `9.27 xx 10^(-28)A m^(2)` |
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Answer» Correct Answer - B Bohr Magnetron `(mu_(l))_("min")=mug=(e)/(4pim_(e))h` Here, `e=1.6xx10^(-19)C, h=6.64xx10^(-34)"J s"` `m_(e)=9.1xx10^(-31)kg` `therefore" "mu_(B)=(1.60xx10^(-19)xx6.64xx10^(-34))/(4xx3.14xx9.1xx10^(-31))=9.27xx10^(-24)" A m"^(2)` |
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| 127. |
A galvanometer having a resistance of `50 Omega`, gives a full scale deflection for a current of 0.05 A. The length (in metres) of a resistance wire of area of cross section `3 xx 10^(-2) cm?` that can be used to convert the galvanometer into an ammeter which can read a maximum of 5 A current is `("Specific resistance of the wirep "= 5 xx 10^(-7)Omega m) `A. 9B. 6C. 3D. 1.5 |
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Answer» Correct Answer - C `S=(I_(g)G)/(I-I_(g)) = (0.05 xx 50)/(5-0.05) = 2.5/4.95 = 50/99 Omega` `therefore S=(rho l)/(A)` or `l = (SA)/rho therefore I=50/99 xx (3 xx 10^(-6))/(5 xx 10^(-7)) = 3.0 m` |
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| 128. |
In a moving coil galvanometer the deflection `(phi)` on the scale by a pointer attached to the spring isA. `((NA)/(kB))I`B. `((N)/(kAB))I`C. `((NAB)/(k))I`D. `((NAB)/(kI))` |
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Answer» Correct Answer - C Since magnetic torque on the coil, `tau=NIAB` This torque is balanced by counter torque `therefore" "kphi=NIAB or phi=((NAB)/(k))I` where k is torsional constant. It is a scalar quantity having dimension of torque or enegy i.e. `[ML^(2)T^(-2)]` |
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| 129. |
The vector form of Biost-Savart law for a current carrying element isA. `d vec(B) = (mu_(0))/(4pi) (idl sin phi)/(r^(2))`B. `d vec(B) = (mu_(0))/(4pi) (ivec(dl) xx hat(r))/(r^(2))`C. `d vec(B) = (mu_(0))/(4pi) (ivec(dl) xx hat(r))/(r^(3))`D. `dvec(B) = (mu_(0))/(4pi) (ivec(dl)xx vec(r))/(r^(2))` |
| Answer» Correct Answer - B | |
| 130. |
A long straight copper wire, of circular cross ection, contains n conduction electrons per unit volume, each of charge q. Show that the current l in the wire is given by `l = nqv pi a^(2)` where v is the drift velocity and a is the radius of the wire. At a radial distance r from the axis of the wire, what is the direction of the magnetic field B due to the current l? Assuming that the magnitude of the field is `B = mu_(0)l//2pir(r ge a)`, obtain an expression for the Lorentz force F on an electron moving with the drift velocity at the surface of the wire. If `l = 10` A and a = 0.5mm, calcualte the magnitude of (a) the drift velocity and (b) the forcce, given that for copper, `n = 8.5 xx 10^(28) m^(-3)` |
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Answer» (a) `9.4 xx 10^(-4)m//s` (b) `6.0 xx 10^(-25) N` |
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| 131. |
Two long straight wires are set parallel to each other Each carries a current in the same directionand the separation between them is 2r. The intensity of the magnetic field midway between them isA. `mu_(0)i//r`B. `4mu_(0)i//r`C. zeroD. `mu_(0)i//4r` |
| Answer» Correct Answer - C | |
| 132. |
Each of two long parallel wires carries a constant current `i` along the same direction . The wires are separated by a distance `2I`. The magnitude of resultant magnetic induction in the symmetric plane of this system located between the wires at a distance `R` from each wire will be |
| Answer» `[(mu_(0)l)/(2pil)]` | |
| 133. |
In a moving coil galvanometer, a rectangular coil of N turns, area of cross-section A and moment of inertia l is suspended in a radial field B through a spring. (a) If a current `i_(0)` produces a deflection of `(pi)/(4)` in the coil, find the torsional constant of the spring (b) Find the maximum deflection surffered by the coil, if a charge Q is passed through it in a short interval of time |
| Answer» `k = (4i_(0)NAB)/(pi), theta_("max") = (Q)/(2) sqrt((NABpi)/(li_(0)))` | |
| 134. |
A circular wire ABC and a straight conductor ADC are carrying current i and are kept in the magnetic field B then considering points A and C A. Force on ABC is more than that on ADCB. Force on ABC is less than that on ADCC. Force on ABC is equal to that on ADCD. Data insufficient |
| Answer» Correct Answer - C | |
| 135. |
A 2.5 m long straight wire having mass of 500 g is suspended in mid air by a uniform horizontal magnetic field B. If a current of 4 A is passing through the wire then the magnitude of the field is `("Take g "10 m s^(-2))`A. 0.5TB. 0.6TC. 0.25TD. 0.8T |
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Answer» Correct Answer - A Here, `m=500g = 0.5kg, I=4A, l=2.5m` As `F=Il B sin 90^(@)," "(because theta=90^(@) and F=mg)` `therefore" "B=(mg)/(Il)=(0.5xx10)/(4xx2.5)=0.5T` |
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| 136. |
A strong magnetic field is applied on a stationary electron, thenA. moves in the direction of the field.B. remained stationary.C. moves perpendicular to the direction of the field.D. moves opposite to the direction of the field. |
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Answer» Correct Answer - B As `vecF=vecq(vecv xx vecB)` As the electron is stationary, `therefore` velocity `vecv=0` `therefore" "vecF=0`. So, electron will remain stationary. |
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| 137. |
A straight wire having mass of 1.2 kg and length of 1m carries a current of 5 A. If the wire is suspended in mid-air by a uniform horizontal magnetic field, then the magnitude of field isA. 0.65TB. 1.53TC. 2.4TD. 3.2T |
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Answer» Correct Answer - C For mid-air suspension the upward force F on wire due to magnetic field B must be balanced by the force due to gravity, then `IlB=mg, B=(mg)/(Il)` Here, `m=1.2kg, g = "10 m s"^(-2), I= 5A, l=1m` `B=(1.2xx10)/(5xx1)=2.4T` |
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| 138. |
A circular coil of radius `R` carries a current `i`. The magnetic field at its centre is `B`. The distance from the centre on the axis of the coil where the magnetic field will be `B//8` isA. `sqrt(2)R`B. `sqrt(3)R`C. `2R`D. `3R` |
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Answer» Correct Answer - 2 `B=(mu_(0)nir^(2))/(2(r^(2)+x^(2))^(3//2))=(1)/(8)xx(mu_(0)ni)/(2r)` |
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| 139. |
Two circular coils are made of two identical wires of same length and carry same current. If the number of turns of the two coils are 4 and 2, then the ratio of magnetic induction at the centres will beA. `2:1`B. `1:2`C. `1:1`D. `4:1` |
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Answer» Correct Answer - 4 `B=(mu_(0)pin^(2)i)/(l)` |
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| 140. |
The magnetic induction at the centre of a current carrying circular coil of radius `10cm` is `5sqrt(5)` times the magnetic induction at a point on its axis. The distance of the point from the centre of the coild in `cm` isA. 5B. 10C. 20D. 25 |
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Answer» Correct Answer - 3 `B=(mu_(0)nir^(2))/(2(r^(2)+x^(2))^(3//2))` |
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| 141. |
A straight wire `(` conductor `)` length `10cm` is kept in a uniform magnetic field of induction `0.02T` . The angle between the conductor and the field direction is `30^(@)`. A current of `5A` is passed through the conductor . Th force on the conductor is `(` in `N )`A. `4xx10^(-3)`B. `5xx10^(-3)`C. `6xx10^(-3)`D. `7xx10^(-3)` |
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Answer» Correct Answer - 2 `F=Bilsintheta` |
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| 142. |
Two circular coils have diameters `10cm` and `20cm` with same number of turns. The ratio of the magnetic field induction produced at the centre of the coils when connected in series isA. `1:2`B. `2:1`C. `4:1`D. `1:4` |
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Answer» Correct Answer - 2 `B alpha (1)/(r)` |
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| 143. |
A circular coil of radius `R` carries an electric current. The magnetic field due to the coil at a point on the axis of the coil located at a distance `r` from the centre of the coil, such that `r gtgt R`, varies asA. `1//r`B. `1//r^(3//2)`C. `1//r^(2)`D. `1//r^(3)` |
| Answer» Correct Answer - D | |
| 144. |
The magnetic field due to a conductor fo unifrom cross section of radius `a` and carrying a steady current is represented byA. B. C. D. |
| Answer» Correct Answer - A | |
| 145. |
A wire carrying current `I` and other carrying `2I` in the sam direction produce a magnetic field `B` at the midpoint. What will be the field when `2I` wire is swiched off?A. `B//2`B. `B`C. `2B`D. `3B` |
| Answer» Correct Answer - 2 | |
| 146. |
A current of `1*0A` flowing in the sides of an equilateral triangle of side `4*5xx10^-2m`. Find the magnetic fied at the centroid of the triangle.A. `4xx10^(-5)T`B. `40T`C. `0.4xx10^(-3)T`D. `4xx10^(-2)T` |
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Answer» Correct Answer - 1 `B=(mu_(0)i)/(4pir)(sintheta_(1)+sintheta_(2))xx3` |
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| 147. |
Two circular coils `P` and `Q` are made from similar wire but radius of `Q` is twice that of `P`. Relation between the values of potential difference across them so that the magnetic induction at their centers may be the same isA. `V_(q)=2V_(p)`B. `V_(q)=3V_(p)`C. `V_(q)=4V_(p)`D. `V_(q)=1//4V_(p)` |
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Answer» Correct Answer - 3 `B=(mu_(0)i)/(2r)` where `i=(V)/(R)` |
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| 148. |
Find the magnetic induction at point `O` if the current carrying wire is in the shape shown in the figure. .A. `(mu_(0)I)/(4pir)`B. `(mu_(0)I)/(4r)+(mu_(0)I)/(2pir)`C. `(mu_(0)I)/(4r)+(mu_(0)I)/(4pir)`D. `(mu_(0)I)/(4r)-(mu_(0)I)/(4pir)` |
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Answer» Correct Answer - 3 `B=B_(1)+B_(2),B=(mu_(0)I)/(4r)+(mu_(0)I)/(4pir)` |
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| 149. |
Find the magnetic induction at point `O` if the current carrying wire is in the shape shown in the figure. .A. `(mu_(0)I)/(2R)`B. `(mu_(0)I)/(2piR)`C. `(mu_(0)I(pi^(2)+4)^(1//2))/(4piR)`D. `(mu_(0)I(pi^(2)+4))/(4piR)` |
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Answer» Correct Answer - 3 `B_(1)=(mu_(0)i)/(4pir)+(mu_(0)i)/(4piR),B_(2)=(mu_(0)i)/(4R),B=sqrt(B_(1)^(2)+B_(2)^(2))` |
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| 150. |
A long solenoid has `200 turns per cm` and carries a current `i`. The magnetic field at its centre is `6.28xx10^(-2) weber//cm^(2)`. Another long soloenoid has `100 turns per cm` and it carries a current `(i)/(3)`. The value of the magnetic field at its centre isA. `1.05X10^(-3)Wb//m^(2)`B. `1.05X10^(-4)Wb//m^(2)`C. `1.05X10^(-2)Wb//m^(2)`D. `1.05X10^(-5)Wb//m^(2)` |
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Answer» Correct Answer - 3 `B=mu_(0)ni` |
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