A magnetic field can be produced by

Question 1. A current i is flowing in a straight conductor of length L. The magnetic field at a point distant

\frac{L}{4}

from its centre will be

4μ0i√5πL
4μ0i2πL
μ0i√2L
Zero

Answer: Option A
: A

By using

B = \frac{μ_{0}}{4 π} . \frac{i}{a} (s i n ϕ_{1} + s i n ϕ_{2})

\Rightarrow B = \frac{μ_{0}}{4 π} . \frac{i}{(L / 4)} (2 s i n ϕ)

Also

s i n ϕ = \frac{L / 2}{\sqrt{5} L / 4} = \frac{2}{\sqrt{5}}

\Rightarrow B = \frac{4 μ_{0} i}{\sqrt{5} π L}

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

π24√2
π28√2
π2√2
π4√2

Answer: Option B
: B

Magnetic field at the centre of circular coil

B_{c i r c u l a r} = \frac{μ_{0}}{4 π} . \frac{2 π i}{r} = \frac{μ_{0}}{4 π} . \frac{4 π^{2} i}{L}

Magnetic field at the centre of square coil

B_{s q u a r e} = \frac{μ_{0}}{4 π} . \frac{8 \sqrt{2} i}{a} = \frac{μ_{0}}{4 π} . \frac{32 \sqrt{2} i}{L}

Hence

\frac{B_{c i r c u l a r}}{B_{s q u a r e}} = \frac{π^{2}}{8 \sqrt{2}}

Question 3. Two infinitely long insulated wires are kept perpendicular to each other. They carry currents

I_{1} = 2 A

and

I_{2} = 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?

2×10−5 T, normally into the plane of the paper, 2√2×10−5 T
2×10−5 T,normally out of the plane of the paper, 2√2×10−5 T
2×10−5 T, normally into the plane of the paper, Zero
2×10−5 T, normally into the plane of the paper, Zero

Answer: Option C
: C

B_{1} = \frac{μ_{0} i_{1}}{2 π r_{1}} = \frac{2 \times 10^{- 7} \times 2}{4 \times 10^{- 2}} = 10^{- 5} B_{2} = \frac{μ_{0} i_{2}}{2 π r_{2}} = \frac{2 \times 10^{- 7} \times 1.5}{3 \times 10^{- 2}} = 10^{- 5} B = B_{1} + B_{2} = 10^{- 5} + 10^{- 5}

= 2 \times 10^{- 5}

, 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

Only inside the rod
Only outside the rod
Both inside and outside the rod
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

μ0NIb
2μ0NIa
μ0NI2(b−a)lnba
μ0IN2(b−a)lnba

Answer: Option C
: C

Number of turns per unit width

= \frac{N}{b - a}

Consider an elemental ring of radius x and with thickness dx Number of turns in the ring

= d N = \frac{N d x}{b - a}

Magnetic field at the center due to the ring element

d B = \frac{μ_{0} (d N) i}{2 x} = \frac{μ_{0} i}{2} . \frac{N d x}{(b - a)} . \frac{1}{x}

∴

Field at the centre

= \int d B = \frac{μ_{0} N I}{2 (b - a)} b \int a \frac{d x}{x}

= \frac{μ_{0} N i}{2 (b - a)} l n \frac{b}{a} .

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

μ04π2ir(π+1)
μ04π.2ir(π−1)
Zero
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

B_{1} = \frac{μ_{0}}{4 π} . \frac{2 i}{r} ⊙

.

The field at o due to circular coil is

B_{2} = \frac{μ_{0}}{4 π} . \frac{2 π i}{r} \otimes

. Both fields will act in the opposite direction, hence the total field at O.
i.e.

B = B_{2} - B_{1} = (\frac{μ_{0}}{4 π}) \times (π - 1) \frac{2 i}{r} = \frac{μ_{0}}{4 π} . \frac{2 i}{r} (π - 1)

	A.	A moving charge
	B.	A changing electric field
	C.	None of these
	D.	Both moving charge & changing electric field

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