12th Grade > Physics
MAGNETIC EFFECTS OF CURRENT MCQs
Magnetic Effects Of Electric Current, Magnetic Effects Of Current(10th And 12th Grade)
Total Questions : 54
| Page 2 of 6 pages
Answer: Option A. -> Soft iron
:
A
Soft iron can be used as the core of an electromagnet because it loses all of its magnetism when the current in the coil is switched off. On the other hand, steel retains its magnetism even after the current is switched off.
Brass and aluminium are non-magnetic materials. Hence, they can not be used as the core of an electromagnet.
:
A
Soft iron can be used as the core of an electromagnet because it loses all of its magnetism when the current in the coil is switched off. On the other hand, steel retains its magnetism even after the current is switched off.
Brass and aluminium are non-magnetic materials. Hence, they can not be used as the core of an electromagnet.
Answer: Option C. -> magnetic field
:
C
According to the right-hand thumb rule, if you are holding a current carrying conductor with the right hand in such a way, that the thumb points in the direction of the current then the direction in which the other fingers encircle the conductor gives the direction of the magnetic field.
:
C
According to the right-hand thumb rule, if you are holding a current carrying conductor with the right hand in such a way, that the thumb points in the direction of the current then the direction in which the other fingers encircle the conductor gives the direction of the magnetic field.
Answer: Option B. -> The magnetic field at any point inside the pipe is zero
:
B
Applying Ampere's law ∮B.dl=μ0i to any closed path inside the pipe, we can understand that as the currect enclosed inside it is zero, the magnetic field will be equal to zero.
:
B
Applying Ampere's law ∮B.dl=μ0i to any closed path inside the pipe, we can understand that as the currect enclosed inside it is zero, the magnetic field will be equal to zero.
Question 15. Two very long, straight and parallel wires carry steady currents I and I respectively. The distance between the wires is d. At a certain instant of time, a point charge q is at a point equidistant from the two wires in the plane of the wires. Its instantaneous velocity v is perpendicular to this plane. The magnitude of the force due to the magnetic field acting on the charge at this instant is
Answer: Option D. -> 0
:
D
According to gives information following figure can be drawn, which shows that direction of magnetic field is along the direction of motion of charge so net force on it is zero.
if the currents are parallel then the net magnetic field at midpoint will be zero, Hence, no magnetic force acts on it.
:
D
According to gives information following figure can be drawn, which shows that direction of magnetic field is along the direction of motion of charge so net force on it is zero.
if the currents are parallel then the net magnetic field at midpoint will be zero, Hence, no magnetic force acts on it.
Answer: Option D. -> (θ∘360∘)μ0i2r
:
D
B=μ0i4πr[θ]
B=(θ∘360)μ0i2r
:
D
B=μ0i4πr[θ]
B=(θ∘360)μ0i2r
Answer: Option A. -> 4×1016
:
A
B=μ0wq4πr;14=4π×10−7w×1.6×10−194π×12×10−10=71.6×1016=4×1016rads−1
:
A
B=μ0wq4πr;14=4π×10−7w×1.6×10−194π×12×10−10=71.6×1016=4×1016rads−1
Answer: Option C. -> 2π×10−5T
:
C
B=μ0i2r[θ2π]=4π10−7×42×3×10−2[3π22π]=2π×10−5T
:
C
B=μ0i2r[θ2π]=4π10−7×42×3×10−2[3π22π]=2π×10−5T
Answer: Option B. -> 1:9
:
B
Bαn2B1B2=[n1n2]2=[13]2B1:B2=1:9
:
B
Bαn2B1B2=[n1n2]2=[13]2B1:B2=1:9
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