Q.1. A thick currentcarrying cable of radius 'R' carries current 'I' uniformly distributed across its crosssection. The variation of magnetic field B(r) due to the cable with the distance 'r' from the axis of the cable is represented by: (2021)
A:
B:
C:
D:
Ans: A
Inside a current carrying cylindrical conductor
Q.2. An infinitely long straight conductor carries a current of 5 A as shown. An electron is moving with a speed of 10^{5 }m/s parallel to the conductor. The perpendicular distance between the electron and the conductor is 20 cm at an instant. Calculate the magnitude of the force experienced by the electron at that instant. (2021)
A: 4π× 10^{–20} N
B: 8 × 10^{–20} N
C: 4 × 10^{–20} N
D: 8π× 10^{–20 }N
Ans: B
f = 8 × 10^{–2}^{0} Newton.
Q.3. Uniform conducting wire of length 12a and resistance 'R' is wound up as a current carrying coil in the shape of,
(i) an equilateral triangle of side 'a'
(ii) a square of side 'a'
The magnetic dipole moments of the coil in each case respectively are: (2021)
A: 3Ia^{2} and 4Ia^{2}
B: 4Ia^{2} and 3Ia^{2}
C: √3Ia^{2 }and 3Ia^{2}
D: 3Ia^{2} and Ia^{2}
Ans: C
(no. of turns = 4)
M2 = a^{2}I × 3 = 3Ia^{2}
(no. of turns = 3)
Q.4. In the product
What will be the complete expression for ? (2021)
A:
B:
C:
D:
Ans: D
Q.5. A spherical conductor of radius 10 cm has a charge of 3.2 × 10^{7} C distributed uniformly. what is the magnitude of electric field at a point 15 cm from the centre of the sphere? (2020)
Ans: D
Q.6. A long sol enoid of 50 cm length having 100 turns carries a current of 2.5 A. The magnetic field at the centre of the solenoid is : (μ_{0} = 4π × 10^{7} Tm A^{1}) (2020)
A: 6.28×10^{5} T
B: 3.14×10^{5} T
C: 6.28×10^{4} T
D: 3.14×10^{4} T
Ans: C
Magnetic field for solenoid is given by
B = μ_{0} nI
n = N/ℓ
= 4π × 10^{7} × 200 × 2.5
= 4 × 3.14 × 10^{–7} × 500
= 2000 × 3.14 × 10^{–7}
= 6.14 × 10^{–4} T
Q.7. A cylindrical conductor of radius R is carrying a constant current. The plot of the magnitude of the magnetic field. B with the distance d from the centre of the conductor, is correctly represented by the figure : (2019)
A:
B:
C:
D:
Ans: C
Solution:
Inside (d < R) Magnetic field inside conductor
or B = Kd ...(i)
Straight line passing through origin
At surface (d = R)
...(ii)
Maximum at surface Outside (d > R)
Q.8. Ionized hydrogen atoms and αparticles with same momenta enters perpendicular to a constant magnetic field, B. The ratio of their radii of their paths r_{H} : r_{a} will be : (2019)
A: 2 : 1
B: 1 : 2
C: 4 : 1
D: 1 : 4
Ans: A
Solution:
Q.9. A 250Turn rectangular coil of length 2.1 cm and width 1.25 cm carries a current of 85 µA and subjected to magnetic field of strength 0.85 T. Work done for rotating the coil by 180º against the torque is: (2017)
A: 4.55 µJ
B: 2.3 µJ
C: 1.15 µJ
D: 9.1 µJ
Ans: D
Solution:
W = MB (cosθ_{1}– cosθ_{2})
When it is rotated by angle 180º then
W = 2MB
W = 2 (NIA)B
= 2 x 250 x 85 x 10^{}^{6}[1.25 x 2.1 x 10^{4}] x 85 x 10^{2}
= 9.1μJ
Q.10. An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current 'I' along the same direction is shown in fig. Magnitude of force per unit length on the middle wire 'B' is given by : (2017)
A:
B:
C:
D:
Ans: C
Solution:
Q.11. A long straight wire of radius a carries a steady current I. The current is uniformly distributed over its crosssection. The ratio of the magnetic fields B and B`, at a radial distances a/2 and 2a respectively, from the axis of the wire is: (2016)
A: 4
B: 1/4
C: 1/2
D: 1
Ans: D
Solution:
Consider two amperian loops of radius a/2 and 2a as shown in the diagram.
Applying Ampere's circuital law for these loops, we get
B_{1} acts at a distance a/2 from the axis of the wire.
Similarly, for bigger amperian loop,
Total current enclosed by amperian loop 2.
at a distance 2a from the axis of the wire.
Q.12. Two identical long conducting wires AOB and COD are placed at right angle to each other, with one above other such that ‘O’ is their common point for the two. The wires carry I_{1} and I_{2} currents, respectively. Point ‘P’ is lying at distance ‘d’ from ‘O’ along a direction perpendicular to the plane containing the wires. The magnetic field at the point ‘P’ will be : (2014)
A:
B:
C:
D:
Ans: B
Solution:
The point P is lying at a distance d along the zaxis.
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