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Magnetics question

2019 · 9 Apr · Shift 1 · Q55
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Magnetics question

2019 · 9 Apr · Shift 1 · Q55

JEE MainPhysicsMagneticsMCQ+4 / −1
A rigid square loop of side 'a' and carrying current I2 is lying on a horizontal surface near a long current I1 carrying wire in the same plane as shown in figure. The net force on the loop due to wire will be : JEE Main 2019 (Online) 9th April Morning Slot Physics - Magnetic Effect of Current Question 164 English
  1. A
    Repulsive and equal to μ\muμ 0I1I2/4 π\piπ
  2. B
    Attractive and equal to μ\muμ 0I1I2/3 π\piπ
  3. C
    Repulsive and equal to μ\muμ 0I1I2/2 π\piπ
  4. D
    Zero
View written solutionFree

Correct answer: A

  1. Magnetic field due to the long straight wire

At a distance rrr from a long straight wire carrying current I1I_1I1​, the magnetic field is

B(r)=μ0I12πr.B(r)=\frac{\mu_0 I_1}{2\pi r}.B(r)=2πrμ0​I1​​.

This field is perpendicular to the plane of the square loop.

  1. Forces on different sides of the square loop

The square loop of side aaa carries current I2I_2I2​ and lies in the same plane as the wire.

From the figure implied by the options, the nearer vertical side of the loop is at distance aaa from the wire, and the farther vertical side is at distance 2a2a2a from the wire.

(i) Horizontal sides

The top and bottom horizontal sides experience equal and opposite vertical forces because the magnetic field depends only on distance from the wire, and corresponding elements on these two sides cancel out.

So, net force due to horizontal sides = 0.

(ii) Vertical sides

Only the two vertical sides contribute to the net horizontal force.

  • Nearer side at distance aaa:
F1=I2a B(a)=I2a⋅μ0I12πa=μ0I1I22π.F_1 = I_2 a\, B(a)= I_2 a \cdot \frac{\mu_0 I_1}{2\pi a} = \frac{\mu_0 I_1 I_2}{2\pi}.F1​=I2​aB(a)=I2​a⋅2πaμ0​I1​​=2πμ0​I1​I2​​.
  • Farther side at distance 2a2a2a:
F2=I2a B(2a)=I2a⋅μ0I12π(2a)=μ0I1I24π.F_2 = I_2 a\, B(2a)= I_2 a \cdot \frac{\mu_0 I_1}{2\pi (2a)} = \frac{\mu_0 I_1 I_2}{4\pi}.F2​=I2​aB(2a)=I2​a⋅2π(2a)μ0​I1​​=4πμ0​I1​I2​​.

These two forces act in opposite directions.

  1. Direction of net force

For the side nearer to the wire, the current in the loop is opposite to the current in the straight wire, so that side is repelled more strongly.

The farther side has current in the same direction as the straight wire, so it is attracted, but less strongly because it is farther away.

Hence the loop experiences a net repulsive force away from the wire.

  1. Net force magnitude
Fnet=F1−F2=μ0I1I22π−μ0I1I24π=μ0I1I24π.F_{\text{net}} = F_1 - F_2 = \frac{\mu_0 I_1 I_2}{2\pi} - \frac{\mu_0 I_1 I_2}{4\pi} = \frac{\mu_0 I_1 I_2}{4\pi}.Fnet​=F1​−F2​=2πμ0​I1​I2​​−4πμ0​I1​I2​​=4πμ0​I1​I2​​.
  1. Matching with options

Thus, the net force on the loop is

μ0I1I24π\boxed{\frac{\mu_0 I_1 I_2}{4\pi}}4πμ0​I1​I2​​​

and it is repulsive.

So the correct option is A.

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