Charges & Magnetism Test

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Charges & Magnetism Test - 6

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Weekly Quiz Competition

Question 1
1 / -0

If we double the radius of a current carrying coil keeping the current unchanged. what happens to the magnetic field at its Centre?

A
halved

B
doubled

C
remains unchanged

D
becomes four times

Solution

As,
B=μ_onI/2a
a ->radius
B ∝1/a
B₁/B₂=a₂/a₁
B₁=2B₂
B₂=(1/2) x B₁,
Magnetic field is halved.
Question 2
1 / -0

A circular arc of metallic wire subtends an angle π/4 at center. If it carries current I and its radius of curvature is r, then the magnetic field at the center of the arc is

A

B

C

D

Solution

We know,
B = μoiπ/4πR
Here,
Substituting the value of theta = π/4
So, Magnetic Field = μoIπ/π(4r x 4)r
=> B = μoI/16r
Question 3
1 / -0

The magnetic field B on the axis of a circular coil at distance x far away from its centre are related as:

A
Bαx^-1

B
Bαx^-4

C
Bαx^-3

D
Bαx^-2

Solution

dB=μ_oidlsin90/4πr²=μ_oidl/4πr²
B_net= ∫dBsinθ
= ∫μ_oidlR/4πr²r== (μ_oiR/4πr³) ∫dl
= (μ_oiR/4πr³)2πR= μ_oiR²/2(R²+x²)^3/2
B= μ_oiR²/2(R2+x²)^3/2
If x>>>R
B= μ_oiR²/2(x²)^3/2

[=B ∝ x^-3]
Question 4
1 / -0

Two concentric coils carry the same current in opposite directions. The diameter of the inner coil is half that of the outer coil. If the magnetic field produced by the outer coil at the common centre are 1 T, the net field at the centre is

A
4T

B
2T

C
1T

D
3T

Solution

B₁=μ₀i/2r,
B2=μ₀i/2.2r=μ₀i/4r,
μ₀i/4r=1T
B0=B₁−B₂=μ₀i/4r=1T
Question 5
1 / -0

The magnetic field due to current element depends upon which of the following factors:

A
Current flowing through it.

B
Distance from it.

C
Its length.

D
All of above.

Solution

From Biot-Savart law, magnetic field at a point p, B= (μ0/4π)∫ [(Idl×r)/ r3]
where r is the distance of point p from conductor and I is the current in the conductor.
Thus magnetic field due to current carrying conductor depends on the current flowing through conductor and distance from the conductor and length of the conductor.
Question 6
1 / -0

A circular coil of radius r carries current I. The magnetic field at its center is B. at what distance from the center on the axis of the coil magnetic field will be B/8

A
√3R

B
√2R

C
2R

D
3R

Solution

B= 2μ₀i⋅πr²/4π(r²+x²)^3/2
Magnetic field at centre= μ₀i/2r
B= μ₀i/2r
Put r= √3
B¹=2μ₀i⋅πr²/4π(4r²)^3/2
=B/8
Hence, √3R distance from the centre magnetic field is equal to magnetic field at centre
Question 7
1 / -0

A circular loop of radius 0.0157 m carries a current of 2 A. The magnetic field at the centre of the loop is

A
3.14 x 10^-5 weber /m²

B
1.57 x 10^-5 weber /m²

C
1.57 x 10^-5 weber /m²

D
8.0 x 10^-5 weber /m²

Solution

The magnetic field due to a circular loop is given by:
B= μ₀2πi/4πr
=10⁻⁷×2π×2/0.0157
=8×10⁻⁵ Wb/m²
Question 8
1 / -0

A straight conductor carrying current I is split into circular loop as shown in figure a , the magnetic induction at the center of the circular loop is

A
Zero

B

C

D
Infinite

Solution

The magnetic field at the center O due to the upper side of the semicircular current loop is equal and opposite to that due to the lower side of the loop.
Question 9
1 / -0

The magnetic field due to circular coil of 200 turns of diameter 0.1m carrying a current of 5A at a point on the axis of the coil at a distance 0.15m from the center of the coil will be

A
3.974 x 10^-5T

B
0.3974 x 10^-5T

C
3974.0 x 10^-5T

D
39.74 x 10^-5T

Solution

B=μ₀2πnIa²/4π (a²+x²)^3/2
=10⁻⁷×2×(22/7)×200×5×(0⋅1/2)²/ [(0⋅1/2)²+(0⋅15)²]^3/2
=39.74x10^-5
Question 10
1 / -0

Wire of length l, carries a steady current I. It is bent first to form a circular coil of one turn. The same wire of same length is now bent more sharply to give two loops of smaller radius the magnetic field at the centre caused by the same current is

A
one third of its initial value

B
nine times of its initial value

C
four times of its initial value

D
unaltered

Solution

Let the radii be r₁ and r₂ respectively.
Since there are two turns of radius r₂, r₁=2r₂
Magnetic field B at the centre of the coil of radius r₁ B₁=μoi/2r₁=μ_oi/4r₂
Magnetic field B at the center of the coil of radius r₂ B₂=2×μ_oi/2r₂
∴ B₂/B₁ =(2× μ_oi/2r₂)/(μ_oi /4r₂) =4
Hence the answer is option C, four times its initial value.