Moving Charges and Magnetism Difficult Questions and Answers | Page - 5

Moving Charges and Magnetism

A current loop consists of two identical semicircular parts each of radius R, one lying in the x-y plane and the other in x-z plane. If the current in the loop is i., the resultant magnetic field due to the two semicircular parts at their common centre is

1. μ₀i
  √2R
1. μ₀i
  2√2R
1. μ₀i
  2R
1. μ₀i
  4R

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Magnetic fields due to the two parts at their common centre are respectively,
B_y = μ₀i and = B_z = μ₀i
4R 4R

Resultant field = √B_y² + B_z
=
μ₀i ² + μ₀i ²
4R 4R

= √2, μ₀i = μ₀i
4R 2√2R

Correct Option: B

Magnetic fields due to the two parts at their common centre are respectively,
B_y = μ₀i and = B_z = μ₀i
4R 4R

Resultant field = √B_y² + B_z
=
μ₀i ² + μ₀i ²
4R 4R

= √2, μ₀i = μ₀i
4R 2√2R

Two circular coils 1 and 2 are made from the same wire but the radius of the 1^st coil is twice that of the 2^nd coil. What potential difference in volts should be applied across them so that the magnetic field at their centres is the same

1. 4
1. 6
1. 2
1. 3
1. None

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If R₁ & R₂ be the radius of the circular wires,R₁/R₂ = 1/2. If same potential is applied on them, current in Ist will be half that in the later. If V potential is applied on them, current in them
= V & V
2R R

Now magnetic field at the centre of circular coil,
= μ₀I
2r

For first wire, field B₁ = μ₀V
2R × 2R

For second wire, field B₂ = μ₀V
2(R/2) × R

Given B₁ = B₂
The given data do not provide any required result. There is a mistake in the framing of the question.

Correct Option: E

If R₁ & R₂ be the radius of the circular wires,R₁/R₂ = 1/2. If same potential is applied on them, current in Ist will be half that in the later. If V potential is applied on them, current in them
= V & V
2R R

Now magnetic field at the centre of circular coil,
= μ₀I
2r

For first wire, field B₁ = μ₀V
2R × 2R

For second wire, field B₂ = μ₀V
2(R/2) × R

Given B₁ = B₂
The given data do not provide any required result. There is a mistake in the framing of the question.

A long solenoid carrying a current produces a magnetic field B along its axis. If the current is double and the number of turns per cm is halved, the new value of the magnetic field is

1. 4B
1. B/2
1. B
1. 2B

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B = μ₀ni
B₁ = (μ₀) n (2i) = (μ₀ni) = B
2

⇒ B₁ = B

Correct Option: C

B = μ₀ni
B₁ = (μ₀) n (2i) = (μ₀ni) = B
2

⇒ B₁ = B

A wire carries a current. Maintaining the same current it is bent first to form a circular plane coil of one turn which produces a magnetic field B at the centre of the coil. The same length is now bent more sharply to give a double loop of smaller radius. The magnetic field at the centre of the double loop, caused by the same current is

1. 4B
1. B/4
1. B/2
1. 2B

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Let I be current and l be the length of the wire.
For Ist case : B = μ₀In = μ₀In × π where 2πr' = l and n = 1
2R l

For IInd Case : l = 2(2πr') ⇒ r' = l
4π

B' = μ₀In = μ₀2I = 4μ₀Iπ = 4B
2r' 2 l l
4π

Correct Option: A

Let I be current and l be the length of the wire.
For Ist case : B = μ₀In = μ₀In × π where 2πr' = l and n = 1
2R l

For IInd Case : l = 2(2πr') ⇒ r' = l
4π

B' = μ₀In = μ₀2I = 4μ₀Iπ = 4B
2r' 2 l l
4π

Two long parallel wires P and Q are both perpendicular to the plane of the paper with distance of 5 m between them. If P and Q carry currents of 2.5 amp and 5 amp respectively in the same direction, then the magnetic field at a point half-way between the wires is

1. 3μ₀
  2π
1. μ₀
  π
1. √3μ₀
  2π
1. μ₀
  π

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When the current flows in both wires in the same direction then magnetic field at half way due to the wire P,
^→_B'_p = μ₀I₁ = μ₀I₁ = μ₀ (where I₁ = 2.5 amp)
2π 5 π5 π2
2

The direction of ⇒_B'_p is downward ⊙

Magnetic field at half way due to wire Q
^→_B'_Q = μ₀I₂ = μ₀ [upward ⊗]
2π 5 π
2

[where I₂ = 2.5 amp]
Net magnetic field at half way
^→_B' = ^→_B'_P + ^→_B'_Q
= - μ₀ + μ₀ = μ₀ (upward)
2π π 2π

Hence, net magnetic field at midpoint = μ₀
2π

Correct Option: D

When the current flows in both wires in the same direction then magnetic field at half way due to the wire P,
^→_B'_p = μ₀I₁ = μ₀I₁ = μ₀ (where I₁ = 2.5 amp)
2π 5 π5 π2
2

The direction of ⇒_B'_p is downward ⊙

Magnetic field at half way due to wire Q
^→_B'_Q = μ₀I₂ = μ₀ [upward ⊗]
2π 5 π
2

[where I₂ = 2.5 amp]
Net magnetic field at half way
^→_B' = ^→_B'_P + ^→_B'_Q
= - μ₀ + μ₀ = μ₀ (upward)
2π π 2π

Hence, net magnetic field at midpoint = μ₀
2π