# MOVING CHARGES AND MAGNETISM

JEE Advanced Physics Syllabus can be referred by the IIT aspirants to get a detailed list of all topics that are important in cracking the entrance examination. JEE Advanced syllabus for Physics has been designed in such a way that it offers very practical and application-based learning to further make it easier for students to understand every concept or topic by correlating it with day-to-day experiences. In comparison to the other two subjects, the syllabus of JEE Advanced for physics is developed in such a way so as to test the deep understanding and application of concepts.

Q1. Two particles A and B of masses mA and mB, respectively, and having the same charge are moving in a plane. A uniform magnetic field exists perpendicular to this plane. The speeds of the particles are vA and vB, respectively and the trajectories are as shown in figure. Then
•   mAvA < mBvB
•  mA vA>mB vB
•  mA<mb
•  mA=mB and vA=vB
Q2.An electron is accelerated from rest through a potential difference V. This electron experiences a force F in a uniform magnetic field. On increasing the potential difference to V', the force experienced by the electron in the same magnetic field becomes 2F. Then, the ratio (V^'/V) is equal to
•  1/4
•  2/1
•  1/2
•  4/1
Q3.  A charged particle moving along +ve x-direction with a velocity v enters a region where there is a uniform magnetic field B_0 (-k ̂ ), from x=0 to x=d. The particle gets deflected at an angle θ from its initial path. The specific charge of the particle is
•   (v cos⁡θ)/Bd
•  (v tan⁡θ)/Bd
•  v/Bd
•  (v sin⁡θ)/Bd
Solution

Q4. A long cylindrical wire of radius ‘a’ carries a current i distributed uniformly over its cross section. If the magnetic fields at distances r<a and R<a from the axis have equal magnitude, then
•  a=(R+r)/2
•  a=√Rr
•  a=Rr/R+r
•  a=R^2/r
Solution
Q5. A Q5.charged particle moves with velocity v =ai ̂+dj ̂ in a magnetic field B =Ai ̂+Dj ̂. The force acting on the particle has magnitude F. Then,

•  F=, if aA=-dD
•  F∝(a^2+b^2 )^(1/2)×(A^2+D^2 )^(1/2)
Q6. A very thin metallic wires placed along X-and Y-axes carry equal currents as shown in figure AB and CD are lines at 45° with the axes. The magnetic fields will be zero on the line
•  AB
•  CD
• Segment OB only of line AB
•  Segment OC only of line CD
Q7.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 X-Y plane and a steady current I flows through the wire. The Z-component of the magnetic field at the centre of the spiral is
•  (μ_0 NI)/(2(b-a)) ln⁡(b/a)
•  (μ_0 NI)/(2(b-a)) ln⁡((b+a)/(b-a))
•  (μ_0 NI)/2b ln⁡(b/a)
•  (μ_0 NI)/2b ln⁡((b+a)/(b-a))
Q8.An electron is projected at an angle θ with a uniform magnetic field. If the pitch of the helical path is equal to its radius, then the angle of projection is
•  tan^(-1)⁡π
•  Ttan^(-1)⁡2π
•  cot^(-1)⁡π
•  cot^(-1)⁡2π
Q9.An infinitely long conductor PQR is bent to form a right angle as shown in figure. A current I flows through PQR. The magnetic field due to this current at the point M is H_1. Now, another infinitely long straight conductor QS is connected to Q so that current is I/2 in QR as well as in QS, the current in PQ remaining unchanged. The magnetic field at M is now H_2. The ratio H_1/H_2 is given by
•  1/2
•  1
•  2/3
•  2
Q10. A small block of mass m, having charge q, is placed on a frictionless inclined plane making an angle θ with the horizontal. There exists a uniform magnetic field B parallel to the inclined plane but perpendicular to the length of spring. If m is slightly pulled on the incline in downward direction, the time period of oscillation will be (assume that the block does not leave contact with the plane)
•  2π√(m/K)
•  2π√(2m/K)
•  2π√(qB/K)
• 2π√(qB/2K)
Solution #### Written by: AUTHORNAME

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