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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. |
A simple pendulum with charge bob is oscillating as shown in the figure. Time period of oscillation is `T` and angular ampliltude is `theta`. If a uniform magnetic field perpendicular to the plane of oscillation is switched on, then A. T will decrease but `theta` will remain constantB. T wil remainconstant but `theta` will decreaseC. Both T and `theta` will remain the sameD. Both T and `theta will decrease |
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Answer» Correct Answer - C Mangetic force is always perpendicular to velocity. So it will always act in radial direction which will change tension at differenct points. But, time period and `theta` will remain unchanged. |
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| 2. |
A proton moving with a constant velocity passes through a region of space without any changing its velocity. If `E` and `B` represent the electric and magnetic fields, respectively. Then, this region of space may haveA. `E=0,B=0`B. `E=0, B!=0`C. `E!=0,B=0`D. `E!=0,B!=0` |
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Answer» Correct Answer - D If both `E` and `B` are zero, the `F_e ` and `F_m` bothare zero,. Hence, velocity may remain constant. Therefore, option a is correct. If `E=0, B!=0` but velocity is parallel or antiparallel to magnetic field, then also `F_e` and `F_m` both are zero. Hence, option b, is also correct. If `E!=0, B!=0 but F_e+F_m=0` then again velocity may remain constant or option d is also correct. |
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| 3. |
In a high tension wire electric current runs from eash to west. Find the direction of magnetic fielsd at points above and below the wire. |
| Answer» When the current flow from east to west, magnetic field lines are circular round it as shown as showin in figure A. And so the magnetic field above the wire is towards north and below the wire towards south. | |
| 4. |
A charged particle of unit mass and unit charge moves with velocity `vecv=(8hati+6hatj)ms^-1` in magnetic field of `vecB=2hatkT`. Choose the correct alternative (s).A. The path of the particle may be `x^2+y^2-4x-21=0`B. The path of the particle may be `x^2+y^2=25`C. The path of the particle may be `y^2+z^2=25`D. The time period of the particle will be 3.14 s |
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Answer» Correct Answer - A::B::D `r=(mV)/(Bq)=((1)(10))/((2)(1))=5m` `T=(2pim)/(Bq)=((2)(pi)(1))/((2)(1))=pi` `=3.14s` Plane of circle is perpendicular to `B`, i.e., `xy`-plane. |
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| 5. |
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))ln(b/a)`B. `(mu_0NI)/(2(b-a))ln ((b+a)/(b-a))`C. `(m_0NI)/(2b)ln(b/a)`D. `(m_0NI)/(2b)((b+a)/(b-a))` |
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Answer» Correct Answer - A a. If we take a small strip of `dr` at distance `r` from centre, then number of turns in this strip would be `dN=(N/(b-a))dr` Magnetic field due to this element at the cenntre of the coil will be `dB=(mu_)(dN)I)/(2r)=(mu_0NI)/((b-a)) (dr)/(2r)` `:. Bint_(r=a)^(r=b)dB=(m_0NI)/(2(b-a))ln(b/a)` `:. current answer is a. |
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| 6. |
Consider a ????cable which consists of an inner wilre of radius `a` surrounded by an outer shell of inner and outer radii `b` and `c` respectively. The inner wire caries a current `I` and outer shell carries an equal and opposite current. The magnetic field at a distance `x` from the axis where `bltxltc` isA. `(mu_0I(c^2-b^2))/(2pix(c^2-a^2))`B. `(mu_0I(c^2-x^2))/(2pix(c^2-a^2))`C. `(mu_0I(c^2-x^2))/(2pix(c^2-b^2))`D. zero |
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Answer» Correct Answer - C `B_x=mu_0/(2pi) I_("in")/x` where `I_("in")=I-[I/(pi(c^2-b^2)][pi(x^2-b^2)]` `=(I(c^2-x^2))/((c^2-b^2))` |
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| 7. |
A circular loop of radius `R` carries current `I_2` in a clockwise direction as shown in figure. The centre of the loop is a distance `D` above a long, straight wire. What are the magnitude and direction of the current `I_1` in the wire if the magnetic field at the centre of loop is zero? |
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Answer» Correct Answer - A::B::D `I_2` produces inwards magnetic field at centre. Hence `I_1` should produce outward magnetic field. or current should be towards right. Further, `(mu_0I_2)/(2R) =((mu_0)/(2pi)) (I_1/D)` `:. I_1=((piD)/R) I_2` |
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| 8. |
A segment `AB` of wire carrying current `I_1` is placed perpendicular to a long straight wire carrying current `I_2` as shown in figure. The magnitude of force experieced by the straight wire `AB` is A. `(mu_0I_1I_2)/(2pi)ln 3`B. `(mu_0I_1I_2)/(2pi) ln 2`C. `(2mu_0I_1I_2)/(2pi)`D. `(mu_0I_1I_2)/(2pi)` |
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Answer» Correct Answer - B At a distance X from current `I_2` `B=mu_0/(2pi)I_2/X` Magnetic force of small element `dX` of wire `AB` `dF=I_2(dX)Bsin90^@` `:. F=int_(x=a)^(x=2a) dF` |
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| 9. |
Two long mutually perpendicular conductors carrying currents `I_1` and `I_2` lie in one plane, Find the locus of points at which the magnetic induction is zero. |
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Answer» Correct Answer - A::B In first quadrant magnetic field ue to `I_1` is outwards and due to `I_2` is inwards. So, net magnetic field may be zero. Similarly, in third quadrant magnetic field due to `I_1` is inwards and due to `I_2` magnetic fiedl is outwards. Hence only in first and third quadrants magnetic field may be zero. Let magnetic field is zero at point P(xy) then `B_(I_1)=B_(I_2)` `:. mu_0/(2pi) I_1/y mu_0/(2pi) I_2/x` `:. y=I_1/I_2x` |
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| 10. |
A conductor consists of a circular loop of radius `R =10 cm` and two straight, long sections as shown in figure. The wire lies in the plane of the paper and carries a current of `i = 7.00 A` Determine the magnitude and direction of the magnetic field at the centre of the loop. |
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Answer» Both straight and circular wire wire produce magnetic fields inwards. `:. B=mu_0/(2pi)i/R+(mu_0i)/(2R)` `=((2xx1^-7)(7))/0.1 +((4pixx10^-7)(7))/(2xx0.1)` `=5.8xx10^-5T` |
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| 11. |
A wire carrying current `i` has the configuration as shown in figure. Two semi-infinite straight sections, both tangent to the same circle, are connected by a circular arc of central angle theta, along the circumference of the circle, with all sections lying in the same plane. What must be for `B` to be zero at the centre of the circle? |
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Answer» Correct Answer - A::B::D Two straight wires produces outward magnetic field by arc of circle produces inward magnetic field. Due to straight wires, `B_1=2[mu_0/(4pi)i/R (sintheta+sin90^[email protected])]` `=mu_0/(2piR) (outwards) `Due to circle arc `B_2=theta/(2pi)((mu_0i)/(2R))`(inward) For net to be zero `B_1=B_2` `or theta=2 rad` |
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| 12. |
A coil with magnetic moment `1.45 A -m^2` is oriented initially with its magnetic moment antiparallel to a uniform `0.835 T` magnetic field. What is the change in potential energy of the coil when it is rotated `180^@` so that its magnetic moment is parallel to the field? |
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Answer» `:./_U=U_(0^@)-U_(180^@)` `=-MBcos09^@+MBCos180^@=-2MB` |
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| 13. |
A conductor carries a constant current `I` along the closed path `abcdefgha` involving `S` of the 12 edges each of length 1. Find the magnetic dipole moment of the closed path. |
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Answer» Correct Answer - A::B Assume equal and opposite currents in wire cf and eh. |
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| 14. |
A conductor `ab` of arbitrary shape carries current I flowing from `b` to `a`. The length vector `ab` is oriented from `a` to `b`. The force `F` experienced by this conductor in as uniform magnetic field B isA. `F=-I(abxxB)`B. `F=I(bxxab)`C. `F=I(baxxB)`D. All of the above |
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Answer» Correct Answer - D `F=I(IxxB)=I(baxxB)` `We can see that all a, b and c optin are same. |
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| 15. |
If as long cylindrical coductor carreis a steady current parallel to its length, thenA. the electric field along the axis is zeroB. the magnetic field along the axis is zeroC. the magnetic field outside the conductor is zeroD. the elctric field outside the conductor is zero |
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Answer» Correct Answer - B::D If current flows in a coductor, then `E!=o` (for inside points) `E=0` (for outside points) `B=mu_0/(2pi)i/R^2` (for inside points) `B=0` at `r=0,` i.e. at centre `B=mu_0/(2pi)i/r` for outside points. |
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| 16. |
A straight long conductor carries along the positive x-axis. Identify the correct statement related to the four points `A(a,a,0), B(a,0,a), C(a,-a,0)` and `D(a,0,-a)`.A. The magnitude of magnetic field at all points is sameB. Fields ast A and B are mutually perpendicularC. Fields at A and C are antiparallelD. All of the above |
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Answer» Correct Answer - D Apply screw law for finding magnetic field around a straight current carrying wire. |
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| 17. |
A long cylidrical conductor of radius `R` carries a current i as shown in figure. The current desity `J` is a function of radius according to `J=br`, where `b` is a constant. Find an expression for the magnetic field `B` a. at a distasnce `r_1ltR` and b.at a distance `r_2gtR,` measured from the axis. |
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Answer» Correct Answer - B::C `i=int_0^r(2pir dr)j=int_0^r(2pir dr)(br)=(2pibr^3)/3` `a. For r_1ltR` `B=mu_0/(2pi) i_m/r_1` `=mu_0/(2pi) ((2pibr_1^3/3)/r_1)` `=(mu_0br_1^2)/3` b. For `r_2ltR` `B=mu0/(2pi) i_m/r_2` `=(mu_0)/(2pi) ((2pibR^3/b)/r_2)=(mu_0bR^30/(3r_2)` |
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| 18. |
A horizontal rod `0.2 m` long is mounted on a balnce and carries a current. At the location of the rod a uniform horizontal magnetic field has magnitude `0.067 T` and direction perpendicular to the rod. The magnetic force on the rod is measured by the balance and is found to be `0.13N`. What is the current? |
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Answer» `F=ilB sin90^@` `:. =F/(lB)=0.13/(0.2xx0.067)` `=9.7A` |
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| 19. |
A cylindrical long wire of radius `R` carries a current `I` uniformly distributed over the cross sectionasl area of the wire. The magnetic field at a distasnce `x` from the surface inside the wire isA. `(mu_0I)/(2pi(R-x)`B. `(mu_0I)/(2pix)`C. `(mu_0I)/(2pi(R+x))`D. none of these |
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Answer» Correct Answer - D From centre `r=(R-x)` `B=mu_0/(2pi) I/R^2r=mu_0/(2pi) 1/R^2(R-x)` |
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| 20. |
A ring of radius R having unifromly distributed charge Q is mounted on a rod suspended by two identical strings. The tension in strings in equilibrium is `T_0.` Now a vertical magnetic field is switched on and ring is rotated at constant angular velocity `omega`. Find the maximum `omega` with which the ring can be rotated if the strings can withstand a maximum tension of `3T_0 //2.` |
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Answer» Correct Answer - B::D In equilibrium, `2T_0=mg` or `T_0=(mg)/2`…….i Magnetic moment `M=iA=(omega/(2pi)Q)(piR^2)` `tau=MBsin90^@=(omegaQR^2)/2` Let `T_1` and `T_2` be the tension in the two strings when magnetic field is switched on `(T_1ltT_2)`. For translation equilibrium of ring in vertical direction. `T_1+T_2=mg`...........i For rotationasl equilibrium `(T_1-T_2)D/2=tau=(omegaBQR^2)/2` or `T_1-T_2=(omegaBQR^2)/2`.........iii Solving eqn ii and iii, we have `T_1=(mg)/2+(omegaBQR^2)/(2D)` As `T_1gtT_2` and maximum values of `T_1` and `(3T_0)/2`, We have `(3T_0)/2=T_0+(omega_maxBQR^2)/(2D)` `:. omega_max=(DT_0)/(BQR^2)` |
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| 21. |
A particle with charge `7.80 muC` is moving with velocity `v = - (3.80 xx 10^3 m/s)hatj`. The magnetic force on the particle is measured to be `F = + (7.60 xx 10^-3 N)hati - (5.20 xx 10^-3 N)hatk` (a) Calculate the components of the magnetic field you can find from this information. (b) Are the components of the magnetic field that are not determined by the measurement of the force? Explain. (c) Calculate the scalar product `B .F`. What is the angle between `B` and `F` |
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Answer» a. `F=q(vxxB)` or `[(7.6xx10^-3)hati-(5.2xx10^-3)hatk]` `=(7.8xx10^-6)[(-3.8xx10^3)hatj(B_xhati+B_yhatj+B_2hatk)]` `:.(7.8xx10^-6)(3.8xx10^-3)(B_x)` `=(-5.2xx10^-3)` `:.B_x=-0.175T` Similarly `(7.8xx10^-6)(-3.8xx10^3)(B_z)` `=7.6x10^-3)` or `B_z=-0.256T` c. From the property of cross product. `F` is always perpendicular to `B`. Hence, `F.B=0` |
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| 22. |
Each of the lettered points at the corners of the cube as shown in Fig. 1.60 represents a positive charge q moving with a velocity of magnitude v in the direction indicated. The region in the figure is in a uniform magnetic field `vec B`, parallel to the x-axis and directed toward right. Copy the figure, find the magnitude and direction of the force on each charge and show the force in your diagram. |
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Answer» Correct Answer - A Apply `F=q(vxxB)` for example let us apply for charged particle at e. `F_e=q[(v/sqrt2hatj-v/sqrt2hatk)xx(Bhati)]` `=(qvB)/sqrt2(-hatj-hatk)` |
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| 23. |
A particle of specific charge `alpha` is projected from origin with velocity `v=v_0hati-v_0hatk` in a uniform magnetic field `B=-B_0hatk`. Find time dependence of velocity and position of the particle. |
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Answer» Velocity of particle at time is `v(t)=v_xhati+v_yhatj+v_zhatk` `=v_0cos(B_0alphat)hati+v_0sin(B_0alphat)hat-v_0hatk` `v_x` and `v_y` can be found in the similar manner as done in Example 2. the position of the particle at time `t` would be `r(t)=xhati+yhatj+zhatk` Here, `z=v_zt=-v_0t` and `x` and `y` are same as in Example 2. Hence, `r(t)=v_0/(B_0alpha)[sin(B_0alphat)hati+1-cos(B_0alphat)hatj]-v_0thatk` |
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| 24. |
Electric field and magnetic fiedl `n` a region of sace is given by `E=E_0hatj` and `B=B_0hatj`. A particle of specific charge alpha is released from origin with velocity `v=v_0hati`. Then path of particelA. is a circleB. is a helix with uniform pitchC. is a helix with non uniform pitchD. is cycloid |
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Answer» Correct Answer - C Electric field (acting along `hatj` direction) will change the velocity component which is parallel to `B=B_0hatj` and `v=v_0hatj` will rotate the particle in a circle. Hence, the net path is helical with variable pitch. |
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| 25. |
An electron and a proton are moving with the same kinetic energy along the same direction, When they pass through a uniform magnetic field perpendicular to the direction of their motion, they describe circular path of the same radius. Is this statement true or false? |
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Answer» `r=sqrt((2km))/(Bq)=rpropsqrtm` `(k,q` and `B` are same) `the path will be a helix. Path is circle when it enters normal to the magnetic field. |
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| 26. |
A charged particle projected in a magnetic field `B=(3hati+4hatj)xx10^-2T` The acceleration of the particle is found to be `a=(xhati+2hatj)m/s^2` find the value of `x` |
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Answer» As we have read e`F_m_|_B` i.e. the acceleration `a_|_B` or `a.b=0` or `(xhati+2hatj).(3hati+4hatj)xx10^=-2=0` `or (3x+8)xx10^-2=0` `x=-8/3m/s^2` |
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| 27. |
When as proton has a velocity `v=(2hati+3hatj)xx10^6m/s`, it experience a force `F=-(1.28xx10^-13hatk)` When its velocity is along the z-axis, it experience a force along the `x`-axis. What is the magnetic field? |
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Answer» Substituting proper values in `F_m=q(vxxB)` We have `-(1.28xx10^-13hatk)=(1.6xx10^-19)[(2hati+3hatj)xx(-B_0hatj)]xx10^6` `:. 1.28=1.6xx2xxB_0` or `B_0=1.28/3.2=0.4` Therefore, the magnetic field is `B=(-0.4hatj)T` |
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| 28. |
A magnetic field of `(4.0xx10^-3hatk)T` exerts a force `(4.0hati+3.0hatj)xx10^-10N` on a particle having a charge `10^-9C` and moving in te `x-y` plane. Find the velocity of the particle. |
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Answer» Given `B=(4xx10^-3hatk)T,q=10^-9C` and magnetic force `F_m=(4.0hati+3.0hatj)xx10^-10N` Let velocity of the particle in `x-y` plane be `v=v_xhati+v_yhatj` Then, from the relation `F_m=q(vxxB)` We have `(4.0hati+3.0hatj)xx10^-10=10^-9[(v_xhati+v_yhatj)xx(4xx10^-3hatk)]` `=(4v_yxx10^-12hati-4v_x xx 10^-12hatj)` Comparing the coefficients of `hati` and `hatj` we have `4xx10^-10=4v_yxx10^-12` `:. v_y=10^2m//s=100m//s` and `3.0xx10^-10=-4v_x xx10^-12` `:. v_x=-75m//s` `:. v=(-75hati+100hatj)m//s` |
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| 29. |
Two particles `A` and `B` of masses `m_A` and `m_B` respectively and having the same charge are moving ina plane. A uniform magnetic field exists perependicular to this plane. The speeds of the particles are `v_A` and `v_B` respectively and the trajectories are as shown in the figure. Then, A. `m_Av_Altm_Bv_B`B. `m_Av_Agtm_Bv_B`C. `m_Altm_B and v_Altv_B`D. `m_A=m_B and v_A=v_B` |
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Answer» Radius of the circle `=(mv)/(Bq)` `or radius propmv ` if `B` and `q` are same. `(Radius)_Agt(Radius)_B` `m_V V_Agtm_Bv_B` `:.` correct option is `b` |
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| 30. |
If the acceleration and velocity of a charged particle moving in a constnt magnetic region is given by `a=a_1hati+a_2hatk, v=b_1hati+b_2hatk.[a_1,a_2,b_1` and `b_2` are constant]. Then choose the wrong statementA. magnetic field may be along y-axisB. `a_1b_1+a_2b_2=0`C. magnetic field is along x-axisD. kinetic energy of particle is always constant |
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Answer» Correct Answer - C a. `B_|_v`, so it may along `y`- axis b. `F_|_v`, `:. a_|_v=0` or `a.v=0` or `a_1b_1=a_2b_2` c. See the logic of option a. d. Magnetic force cannot change the kinetic energy of a particle. |
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| 31. |
A particle having mass m and charge `q` is released from the origin in a region in which ele field and magnetic field are given by `B=-B_0hatj`and `E=E_0hatk`. Find the y- component of the velocity and the speed of the particle as a function of it z-coordinate. |
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Answer» Correct Answer - B work is done only by electrostatic force. Hence, from work energy theorem. `=1/2mv^2=` work done by electrostatic force only `=(qE_0)z` or speed `v=sqrt((2qE_0z)/m)` particle rotates in a plane perpendicular in B. i.e. in xy-plane only. Hence `v_y=0` |
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| 32. |
A charged particle having charge `10^-6` C and mass of `10^-10` kg is fired from the middle of the plate making an angle `30^@` with plane of the plate. Length of the plate is `0.17 m` and it is separated by `0.1 m`. Electric field `E = 10^-3N/C` is present between the plates. Just outside the plates magnetic field is present. Find the velocity of projection of charged particle and magnitude of the magnetic field perpendicular to the plane of the figure, if it has to graze the plate at C and A parallel to the surface of the plate. (Neglect gravity) |
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Answer» To graze at `C` using equation of trajectory of parabola, `y=xtantheta-(ax^2)/(2v^2cos^2theta)` .(i) Her, `a=(qE)/m=(10^-6xx10^-3)/10^-10` `=10m//s^2` Substituting in eqn i we have `0.05=0.17tan30^@-(10xx(0.17)^2)/(2v^2xx(sqrt3//2)^2)` Solving this equation we have `v=2m//s` In magnetic field `AC=2r` or `0.1=2r` or `r=0.05m=(mvcos30^@)/(Bq)` `:. B=(mvcos30^@)/((0.05)q)` `=((10^-10)(2)(sqrt3//2))/((0.05)(10^-6))` `=3.46xx10^-3T` `=3.46mT` |
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| 33. |
A proton, a deuteron and an `alpha`- particle having the same kinetic energy are moving in circular trajectors in a constant magnetic field. If `r_p, r_d` and `r_(alpha)` denote respectively the radii of the trajectories of these particles thenA. `r_(alpha)=r_pltr_d`B. `r_(alpha)gtr_dgtr_p`C. `r_(alpha)=r_dgtr_p`D. `r_p=r_d=r_(alpha)` |
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Answer» Radius of the circular path is given by `r=(mv)/(Bq)=sqrt((2Km))/(Bq)` Here `K` is the kinetic energy to the particle, Therefore, `rpropsqrtm/q` if `K` and `B` are same `r_p:r_d=r_(alpha)=sqrt1/1:sqrt2/1:sqrt4/1=1:sqrt2:1` Hence, `r_(alpha)=r_pltr_d` `:.` correct option is a. |
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| 34. |
Can a charged particle be accelerated by a magnetic field. Can its speed be increased? |
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Answer» Magnetic force may be non zero. Hence, acceleration due to magnetic force may be non zero. magnetic force is always perpendicular to velocity. Hence, its power is always zero or work done by magnetic force is always zero. Hence, it can be change the speed of charged particle. |
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| 35. |
In the relation `F=q(VxxB),` which paiers are always perpendicular to each other. |
| Answer» From the property of cross produce `F` is always perpendicular to both `v` and `B`. | |
| 36. |
A charge `q=4muC` has as instantaneous velociyt `v=(2hati-3hatj+hatk)xx10^6m/s` in a uniform magnetic field `B=(2hati+5hatj-3hatk)xx10^-2T`. What is the force on the charge? |
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Answer» `F=q(vxxB)` here `q` has to be substituted with sign. |
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| 37. |
A deuteron (the nucleus of an isotope of hydrogen) has a mass of `3.34 xx 10^-27` kg and a charge of `+e`. The deuteron travels in a circular path with a radius of `6.96 mm` in a magnetic field with magnitude `2.50 T`.(a) Find the speed of the deuteron. (b) Find the time required for it to make half of a revolution. (c) Through what potential difference would the deuteron have to be accelerated to acquire this speed? |
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Answer» Correct Answer - B a. `r=(mv)/(Bq)impliesv=(Bqr)/m` b. `t=T/2=(pim)/(Bq)` c. `r=sqrt((2qVm))/(Bq)` `:. V=(r^2B^2q^2)/(2qm) =(r^2B^2q)/(2m)` |
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| 38. |
An electron beam projected along positive x-axis deflects along the positive y-axis. If this deflection is caused by a magnetic field, what is the direction of the field? |
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Answer» `F=q(vxxB)` `F` is along positin y-direction `q` is negative and `v` is along positive x-direction. Therefore, `B` should be along positive z-direction. |
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| 39. |
If a beam of electrons travels in a straight line in a certain region. Can we say there is no magnetic field? |
| Answer» May be possible that `theta=0^@` or `180^@` betwene `v` and `B`, so that `F_m=0` | |
| 40. |
In a certain region, uniform electric field `E=-E_0hatk` and magnetic field `B = B_0hatk` are present. At time `t = 0`, a particle of mass in and charge q is given a velocity `v = v-0hatj + v-0hatk`. Find the minimum speed of the particle and the time when it happens so. |
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Answer» `a_e=(qE_0)/m` (along negative z-direction) Electric fiedl will make z - component of velocity zero. At that time speed of the particle will be minimum and that minimum, speed ils the other component i.e. `v_0`. `v_z=u_z+a_zt` `or 0=v_0 -(qE_0)/mt` `or t=(mv_0)/(qE_0)` |
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| 41. |
Assertion: A charged particle is moving in a circle with constant speed in uniform magnetic field. If we increase the speed of particle to twice, its acceleration will become four times. Reason: In circular path of radius `R` with constant speed `v`, acceleration is given by `v^2/R`A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of AssertionC. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true. |
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Answer» Correct Answer - D `Rpropv,` by increasing the speed two times radius also becomes two times. Hence, acceleration `(=v^2/R)` will also become only two times. |
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| 42. |
A charged particle moves in a circular path in a uniform magnetic field. If its speed is reduced, then its tiem period willA. increaseB. decreaseC. remain sameD. none of these |
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Answer» Correct Answer - C `T=(2m)/(Bq)`, independent of v. |
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| 43. |
a charged particle moves in a gravity free space without change in velocity. Which of the following is/are possible?A. `E=0 and B!=0`B. `E!=0 and B=0`C. `E!=0 and B!=0`D. `E=0, B=0` |
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Answer» Correct Answer - A::C::D `a. v` is parallel o anti paralle to B c. `qE+q(vxxB)=0` or `E=-(vxxB)=(Bxxv)` |
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| 44. |
A rectangular coil of area `5.0xx10^-4m^2` and `60` turns is pivoted about one of its vertical sides. The coil is in a radial horizontal magnetic field of `9xx10^-3T`. What is the torsional constant of the spring connected to the coil if a current of `0.20mA` produces an angular diflection of `18^@`? |
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Answer» From the equation, `i=(k/(NAB))phi` We find that torsional constant of the spring is given by `k=(NABi)/phi` Substituting the values is `SI` units we have `k=((60)(5.0xx10^-4)(9xx10^-3)(0.2xx10^-3))/18` `=3xx10^-9N-m//degree` |
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| 45. |
Consider the current carrying loop shown in figure formed of radial lines and segments of circles whose centres are at point `P`. Find the magnitude and direction of B at point P. |
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Answer» Magnetic field at `P` due to straight wires `=0` Due to circular wires one is outwards (of radius a) and other is inwards`. 60^@` means `1/6` th of whole circle `:.B=1/6[(mu_0i)/(2a)-(mu_0i)/(2b)]` (outwards) |
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| 46. |
A closely wound, circular coil with radius `2.40 cm` has 800 turns. (a) What must the current in the coil be if the magnetic field at the centre of the coil is`0.0580 T`? (b) At what distance `x` from the centre of the coil, on the axis of the coil, is the magnetic field half its value at the centre? |
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Answer» Correct Answer - A::D a. `B-1=(mu_0Ni)/(2R)` `b. B_2 =(mu_0NiR^2)/(2(R^2+x^2)^(3/2)) given B_2=B_1/2` |
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| 47. |
An electron moving in a circular orbit of radius `R` with frequency `f`. The magnetic field at the centre of the orbit isA. `(mu_0ef)/(2piR)`B. `(mu_0ef)/(2R)`C. `(muef^2)/(2R)`D. zero |
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Answer» Correct Answer - B Equivalent current `ii=qf=ef` `B=(mu_0i)/(2R)=(mu_0ef)/(2R)` |
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| 48. |
The figure shows a point `PO` on the axis of a circular loop carrying current `I`. The corect direction of magnetic field vector at `P` due to `dI` is respectively by A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - A `dB=mu_0?(4pi)i/r^3(dIxxr)` `:.` dB is in the direction `Ixxr`. Hence dI towards P. |
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| 49. |
`abcd` is a square. There is a current `I` in wire `efg` as shown. Choose the correct options. A. Net magnetic field at a in inwardsB. Net magnetic field at b is zeroC. Net magnetic field at c is outwardsD. Net magnetic field at d is inwards |
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Answer» Correct Answer - A::C::D a. Point as lies to the right hand side of `ef ` and `fg`. Hence, both wires prodce inward magnetic field. Hence net magnetic field is inwards. Same logic can be applied for other points also. |
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| 50. |
A particle having charge of `20muC` and mass `20mug` moves along as circle of radius `5cm` under the action of a magnetif field `B=0.1` tesla. When the particle is at `P`, uniform transverse electric field is switched on an it is found that the particle continues along the tangent with a uniform velocity. Find the electric field A. `2V/m`B. `0.5V/m`C. `5V/m`D. `1.5V/m` |
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Answer» Correct Answer - B `r=(mv)/(Bq)` (durign circular path) `:. v=(Bqr)/m` `now, qE=(B^2qr)/m` `:. E=((0.1)^2(20xx10^-6)(5xx10^-2))/((20xx10^-9))` `=0.5V/m` |
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