12th Grade > Physics
MAGNETIC EFFECTS OF CURRENT MCQs
Magnetic Effects Of Electric Current, Magnetic Effects Of Current(10th And 12th Grade)
Total Questions : 54
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Answer: Option B. -> μ04π.2ir(π−1)
:
B
The given shape is equivalent to the following diagram
The field at o due to straight part of conductor is B1=μ04π.2ir⊙.
The field at o due to circular coil isB2=μ04π.2πir⊗. Both fields will act in the opposite direction, hence the total field at O.
i.e. B=B2−B1=(μ04π)×(π−1)2ir=μ04π.2ir(π−1)
:
B
The given shape is equivalent to the following diagram
The field at o due to straight part of conductor is B1=μ04π.2ir⊙.
The field at o due to circular coil isB2=μ04π.2πir⊗. Both fields will act in the opposite direction, hence the total field at O.
i.e. B=B2−B1=(μ04π)×(π−1)2ir=μ04π.2ir(π−1)
Answer: Option D. -> Hit the wall on the left side of the shooter.
:
D
The answer can be found out using Fleming's left-hand rule:
According to the order:Thumb - Force, Forefinger is Magnetic Field, Middle finger- Current. The direction of current is always opposite to the direction of flow of electrons.
The electrons will move towards the wall which is on the left side of the shooter.
:
D
The answer can be found out using Fleming's left-hand rule:
According to the order:Thumb - Force, Forefinger is Magnetic Field, Middle finger- Current. The direction of current is always opposite to the direction of flow of electrons.
The electrons will move towards the wall which is on the left side of the shooter.
Answer: Option C. -> 8×10−5N
:
C
F=μ0i,i2l2πr=2×10−7×10×2×210×10−2F=8×10−5N
:
C
F=μ0i,i2l2πr=2×10−7×10×2×210×10−2F=8×10−5N
Answer: Option C. -> 7.2×10−4N attraction
:
C
F=μ0i1i2l2π[1r1−1r2]=2×10−7×20×10×0.2[110−2−110×10−2]=8×10−6×[910]×102=7.2×10−4N
:
C
F=μ0i1i2l2π[1r1−1r2]=2×10−7×20×10×0.2[110−2−110×10−2]=8×10−6×[910]×102=7.2×10−4N
Answer: Option C. -> μ0NI2(b−a)lnba
:
C
Number of turns per unit width =Nb−a
Consider an elemental ring of radius x and with thickness dx Number of turns in the ring =dN=Ndxb−a
Magnetic field at the center due to the ring element
dB=μ0(dN)i2x=μ0i2.Ndx(b−a).1x
∴ Field at the centre
=∫dB=μ0NI2(b−a)b∫adxx
=μ0Ni2(b−a)lnba.
:
C
Number of turns per unit width =Nb−a
Consider an elemental ring of radius x and with thickness dx Number of turns in the ring =dN=Ndxb−a
Magnetic field at the center due to the ring element
dB=μ0(dN)i2x=μ0i2.Ndx(b−a).1x
∴ Field at the centre
=∫dB=μ0NI2(b−a)b∫adxx
=μ0Ni2(b−a)lnba.
Answer: Option C. -> Both inside and outside the rod
:
C
Magnetic field lies inside as well as outside the solid current carrying conductor.
:
C
Magnetic field lies inside as well as outside the solid current carrying conductor.
Answer: Option D. -> Both moving charge & changing electric field
:
D
A moving charge and changing electric field both produces magnetic field.
:
D
A moving charge and changing electric field both produces magnetic field.
Question 10. In the figure given below, a coil is placed between two permanent magnets and it is rotated with the help of an axle on which two rings R1 and R2 are internally attached. B1 and B2 represent metallic brushes. G represents galvanometer.
As the coil starts to rotate, the current flows through ABCD as shown. The coil is rotating in:
As the coil starts to rotate, the current flows through ABCD as shown. The coil is rotating in:
Answer: Option A. -> Clockwise direction
:
A
Using Fleming's right-hand rule, point your index finger of your right hand towards the direction of the magnetic field i.e. from north to south. Middle finger towards the direction of the current. You can now see that the thumb points upwards which tell that the coil rotates in a clockwise direction.
:
A
Using Fleming's right-hand rule, point your index finger of your right hand towards the direction of the magnetic field i.e. from north to south. Middle finger towards the direction of the current. You can now see that the thumb points upwards which tell that the coil rotates in a clockwise direction.