Magnetic Field Due to Circular Current Carrying Wire - NEET Physics Questions
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Magnetic Field Due to Circular Current Carrying Wire

Question 11: moderate

A wire is bent in the form of a circular arc with a straight portion AB. Magnetic induction at O
when current I is flowing in the wire, is :

1. \[\frac{\mu_{0}}{2r}\left( \pi-\theta+tan\theta \right)\]
2. \[\frac{\mu_{0}I}{2\pi r}\left( \pi+\theta-tan\theta \right)\]
3. \[\frac{\mu_{0}I}{2\pi r}\left( \pi-\theta+tan\theta \right)\]
4. \[\frac{\mu_{0}I}{2\pi r}\left( -tan\theta+\pi-\theta \right)\]
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Question 12: moderate

A wire carrying current I has the configuration as shown in Fig. Two semi-infinite straight
sections both tangent to the same circle are connected by a circular arc of central angle θ,
along the circumference of the circle, with all sections lying in the same plane. If the magnetic
field at the centre of the circle is zero, then θ is:

1. π/2 rad
2. 2/π rad
3. 2 rad
4. 1 rad
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Question 13: moderate

Radius of the current carrying coil is R. If magnetic field at any point on the axis of the coil is \[Bx=\frac{B_{0}}{64}\] then find out axial distance of this point :

1. √3 R
2. √5 R
3. √15 R
4. R
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Question 14: moderate

Three long, straight and parallel wires carrying currents are arranged as shown in figure. The force experienced by 10 cm length of wire Q is :

1. \[1.4\times 10^{-4}N towards the right\]
2. \[1.4\times 10^{-4}N towards left\]
3. \[2.6\times 10^{-4}N to the right\]
4. \[2.6\times 10^{-4}N to the left\]
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Question 15: moderate

Same current i = 2A is flowing in a wire frame as shown in figure. The frame is a combination of two equilateral triangles ACD and CDE of side 1m. It is placed in uniform magnetic field B = 4T acting perpendicular to the plane of frame. The magnitude of magnetic force acting on the frame is :

1. 24 N
2. Zero
3. 16 N
4. 8 N
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Question 16: moderate

A helium nucleus is moving in a circular path of radius \(0.8\text{ m}\). If it takes \(2\text{ sec}\) to complete one revolution, the magnetic field produced at the centre of the circle is:

1. \(\mu_0 \times 10^{-19}\text{ T}\)
2. \(\frac{10^{-19}}{\mu_0}\text{ T}\)
3. \(2 \times 10^{-19}\text{ T}\)
4. \(\frac{2 \times 10^{-19}}{\mu_0}\text{ T}\)
View Answer

Current is \(I = \frac{q}{T} = \frac{2e}{2} = e = 1.6 \times 10^{-19}\text{ A}\). The magnetic field at the centre is \(B = \frac{\mu_0 I}{2R} = \frac{\mu_0 (1.6 \times 10^{-19})}{2(0.8)} = \mu_0 \times 10^{-19}\text{ T}\).