## Lakshya Education MCQs

Question: A magnetic field can be produced by
Options:
 A. A moving charge B. A changing electric field C. None of these D. Both moving charge & changing electric field
Answer: Option D
: D

A moving charge and changing electric field both produces magnetic field.

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## More Questions on This Topic :

Question 1. A current i is flowing in a straight conductor of length L. The magnetic field at a point distant L4 from its centre will be
1.    4μ0i√5πL
2.    4μ0i2πL
3.    μ0i√2L
4.    Zero
Answer: Option A
: A

By using B=μ04π.ia(sinϕ1+sinϕ2)
B=μ04π.i(L/4)(2sinϕ)
Also sinϕ=L/25L/4=25
B=4μ0i5πL
Question 2. The ratio of the magnetic field at the centre of a current carrying circular wire and the magnetic field at the centre of a square coil made from the same length of wire will be
1.    π24√2
2.    π28√2
3.    π2√2
4.    π4√2
Answer: Option B
: B

Magnetic field at the centre of circular coil Bcircular=μ04π.2πir=μ04π.4π2iL
Magnetic field at the centre of square coil
Bsquare=μ04π.82ia=μ04π.322iL
Hence BcircularBsquare=π282
Question 3. Two infinitely long insulated wires are kept perpendicular to each other. They carry currents I1=2 A and I2=1.5 A.

i) Find the magnitude and direction of the magnetic field at P.

ii) If the direction of current is reversed in one of the wires. What would be the magnitude of the field B?

1.    2×10−5 T, normally into the plane of the paper, 2√2×10−5 T
2.    2×10−5 T,normally out of the plane of the paper, 2√2×10−5 T
3.    2×10−5 T, normally into the plane of the paper, Zero
4.    2×10−5 T, normally into the plane of the paper, Zero
Answer: Option C
: C

B1=μ0i12πr1=2×107×24×102=105B2=μ0i22πr2=2×107×1.53×102=105B=B1+B2=105+105
=2×105, normally in to the place of the paper, zero
Question 4. If a copper rod carries a direct current, the magnetic field associated with the current will be
1.    Only inside the rod
2.    Only outside the rod
3.    Both inside and outside the rod
4.    Neither inside nor outside the rod
Answer: Option C
: C

Magnetic field lies inside as well as outside the solid current carrying conductor.
Question 5. A coil having N turns is wound tightly in the form of a spiral with inner and outer radii a and b respectively. When a current I passes through the coil, the magnetic field at the centre is
1.    μ0NIb
2.    2μ0NIa
3.    μ0NI2(b−a)lnba
4.    μ0IN2(b−a)lnba
Answer: Option C
: C

Number of turns per unit width =Nba
Consider an elemental ring of radius x and with thickness dx Number of turns in the ring =dN=Ndxba
Magnetic field at the center due to the ring element
dB=μ0(dN)i2x=μ0i2.Ndx(ba).1x
Field at the centre
=dB=μ0NI2(ba)badxx
=μ0Ni2(ba)lnba.
Question 6. An infinitely long straight conductor is bent into the shape as shown in the figure. It carries a current of i ampere and the radius of the circular loop is r metre. Then the magnetic induction at its centre will be

1.    μ04π2ir(π+1)
2.    μ04π.2ir(π−1)
3.    Zero
4.    Infinite
Answer: Option B
: 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=B2B1=(μ04π)×(π1)2ir=μ04π.2ir(π1)

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