A straight current-carrying wire carrying current \(I\) passes perpendicular to the plane of an imaginary rectangular loop \(PQRS\), passing through its centre \(O\) (into the diagram). The diagonals intersect at \(60^\circ,\) and side \(PS\) is smaller than side \(PQ\). The value of \(\int \vec{B} \cdot d\vec{l}\) evaluated from \(P\) to \(Q\) (along \(PQ\)) has the magnitude:
1. \(\dfrac{\mu_{0} I}{6}\) 2. \(\dfrac{2 \mu_{0} I}{6}\)
3. \(\dfrac{4\mu_{0} I}{6}\) 4. \(\dfrac{5\mu_{0} I}{6}\)

Subtopic:  Ampere Circuital Law |
 52%
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Two small current-carrying loops carrying currents in the clockwise direction are placed in the same plane, separated by a distance \(d\) (which is much larger than the size of the loops). The two loops:
1. attract each other.
2. repel each other. 
3. exert no force on each other, but exert a torque.
4. neither exert any force nor any torque on each other.
Subtopic:  Current Carrying Loop: Force & Torque |
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A long current-carrying solenoid produces a magnetic field \(B\) at its centre, \(O\). When a current-carrying wire is placed parallel to the axis of the solenoid, the field at \(O\) has the magnitude \(2B\). The field due to wire has the magnitude (at \(O\)) of:
1. \(B\) 2. \(3B\)
3. \(\dfrac {B} {\sqrt3}\) 4. \(\sqrt 3~ B\)
Subtopic:  Magnetic Field due to various cases |
 57%
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A wire carrying a current \(i\) is bent into the form of an arc of a circle with center \(O\), joined smoothly to two long, straight wires at its ends. The magnetic field at the centre \(O\) is twice that due to the straight portions. The angle subtended at the centre \(O\) by the arc is:
1. \(\theta=\dfrac{\pi}{3}\) 2. \(\theta=\dfrac{\pi}{2}\)
3.  \(\theta=1\) rad 4. \(\theta=2\) rad
Subtopic:  Magnetic Field due to various cases |
 62%
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The magnetic field at a point (\(P\)) on the axis of a circular current carrying wire is \(\dfrac18\) of the field at its centre. The radius of the circular curve is \(R.\) The distance between \(P\) and the cente of the circle \((OP).\) is:

Then,
1. \(OP=R\) 2. \(OP=\dfrac R2\)
3. \(OP=\sqrt3R\) 4. \(OP=8R\)
Subtopic:  Magnetic Field due to various cases |
 76%
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The magnetic field at the center \(O\) of the semi-circular part of the current-carrying wire, due to the curved and the straight wires (very long), is:
1. \({\Large\frac{\mu_0i}{4R}}\)
2. \({\Large\frac{\mu_0i}{4R}}+{\Large\frac{\mu_0i}{2\pi R}}\)
3. \({\Large\Big(\frac{\mu_0i}{4R}+\frac{\mu_0i}{4\pi R}\Big)} \)
4. \({\Large\Big[\Big(\frac{\mu_0i}{4R}\Big)^2+\Big(\frac{\mu_0i}{2\pi R}\Big)^2\Big]^{1/2}} \)
Subtopic:  Magnetic Field due to various cases |
 67%
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The magnitude of the integral of the quantity \(\int\vec B\cdot d\vec{ l}\) around the loop \(PQR\) of the equilateral triangle is \(K.\) The field at the centre of the long solenoid is:
              
1. \(\dfrac{K}{a}\) 2. \(\dfrac{K}{b}\)
3. \(\dfrac{K}{a-b}\) 4. \(\dfrac{K}{a+b}\)
Subtopic:  Ampere Circuital Law |
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A current \(i\) flows through a semi-circular loop of radius \(r,\) attached to two long straight wires along the open diameter of the loop. The magnetic field at the centre of the loop is:
1. \(\dfrac{\mu_0i}{4r}\)
2. \(\dfrac{\mu_0i}{4r}+\dfrac{\mu_0i}{2\pi r}\)
3. \(\dfrac{\mu_0i}{4r}+\dfrac{\mu_0i}{4\pi r}\)
4. \(\left[\left(\dfrac{\mu_0i}{4r}\right)^2+\left(\dfrac{\mu_0i}{4\pi r}\right)^2\right]^{\frac12} \)
Subtopic:  Magnetic Field due to various cases |
 84%
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A long thin wire of length \(L\) carrying a current \(i,\) is wrapped uniformly around a long solenoid of volume \(V.\) The radius of the solenoid is \(r.\) The magnetic field at its centre is:
1. \(\dfrac{\mu_0~L~r~i}{V}\) 2. \(\dfrac{\mu_0~L~r~i}{2\pi~V}\)
3. \(\dfrac{\mu_0~L~r~i}{2V}\) 4. \(\dfrac{\mu_0~L~r~i}{4\pi~V}\)
Subtopic:  Magnetic Field due to various cases |
 51%
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Two semi-circular current-carrying wires are placed in the same plane so that they share a common centre. The magnetic field due to any one of them at the common centre has the magnitude, \(B_O\). When one of them is tilted so that it is in a perpendicular plane, with the same centre, the magnetic field at the common centre is \(B\). Then,
1. \(B =2B_O\)
2. \(B~=\dfrac{B_O}{2}\)
3. \(B=\sqrt 2 B_O\)
4. \(B=\dfrac{B_O}{\sqrt 2}\)
Subtopic:  Magnetic Field due to various cases |
 74%
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