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A current i ampere flows in a circular arc of wire whose radius is R, which subtend an angle $\frac{3 π}{2}$ radian at its centre. The magnetic induction B at the centre is :

$\frac{μ_{0} i}{R}$

$\frac{μ_{0} i}{2 R}$

$\frac{2 μ_{0} i}{R}$

$\frac{3 μ_{0} i}{8 R_{}}$

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Explanation

(d) $B = \frac{μ_{0}}{4 π} \frac{(2 π - θ) i}{R} = \frac{μ_{0}}{4 π} \frac{(2 π - \frac{π}{2}) \times i}{R} = \frac{3 μ_{0} i}{8 R}$

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The magnetic induction at a point P which is distant 4 cm from a long current carrying wire is $10^{- 8}$ Tesla . The field of induction at a distance 12 cm from the same current would be :

$3.33 \times 10^{- 9}$ Tesla

1.11 $\times 10^{- 4}$ Tesla

$3 \times 10^{- 3}$ Tesla

$9 \times 10^{- 2}$ Tesla

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Explanation

(a) $B = \frac{μ_{0}}{4 π} \cdot \frac{2 i}{r} \Rightarrow \frac{B_{1}}{B_{2}} = \frac{r_{2}}{r_{1}} \Rightarrow \frac{10^{- 8}}{B_{2}} = \frac{12}{4} \Rightarrow B_{2} = 3.33 \times 10^{- 9}$ Tesla

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Two straight horizontal parallel wires are carrying the same current in the same direction, d is the distance between the wires. You are provided with a small freely suspended magnetic needle. At which of the following positions will the orientation of the needle be independent of the magnitude of the current in the wires

At a distance d/2 from any of the wires in any plane

At a distance d/2 from any of the wires in the horizontal plane

Anywhere on the circumference of a vertical circle of radius d and centre halfway between the wires

At points halfway between the wires in the horizontal plane

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Explanation

(d)

At a distance d/2 from any of the wires, the orientation of the needle be independent of the magnitude of current in the wires; since the Lorentz force acting due to the two wires will be equal and in opposite direction.

At these points, the resultant field = 0

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A circular coil of radius R carries an electric current. The magnetic field due to the coil at a point on the axis of the coil located at a distance r from the centre of the coil, such that r >> R, varies as

$\frac{1}{r}$

$\frac{1}{r^{3 / 2}}$

$\frac{1}{r^{2}}$

$\frac{1}{r^{3}}$

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Explanation

(d) $B = \frac{μ_{0}}{4 π} . \frac{2 {πNiR}^{2}}{r^{3}} \Rightarrow B \propto \frac{1}{r^{3}}$

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The magnetic induction due to an infinitely long straight wire carrying a current i at a distance r from the wire is given by

$|B| = (\frac{μ_{0}}{4 π}) \frac{2 i}{r}$

$|B| = (\frac{μ_{0}}{4 π}) \frac{r}{2 i}$

$|B| = (\frac{4 π}{μ_{0}}) \frac{2 i}{r}$

$|B| = (\frac{4 π}{μ_{0}}) \frac{r}{2 i}$

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Explanation

(a)

$T h e m a g n e t i c i n d u c t i o n d u e t o a n i n f i n i t e l y l o n g s t r a i g h t w i r e = |B| = (\frac{μ_{0}}{4 π}) \frac{2 i}{r}$

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Two concentric circular coils of ten turns each are situated in the same plane. Their radii are 20 and 40 cm and they carry respectively 0.2 and 0.3 ampere current in opposite direction. The magnetic field in $w e b e r / m^{2}$ at the centre is :

$\frac{35}{4} μ_{0}$

$\frac{μ_{0}}{80}$

$\frac{7}{80} μ_{0}$

$\frac{5}{4} μ_{0}$

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Explanation

(d) Two coils carrying current in opposite direction, hence net magnetic field at centre will be difference of the two fields.
i.e.

$B_{n e t} = \frac{μ_{0}}{2} \cdot N [\frac{i_{1}}{r_{1}} - \frac{i_{2}}{r_{2}}] = \frac{10 μ_{0}}{2} [\frac{0.2}{0.2} - \frac{0.3}{0.4}] = \frac{5}{4} μ_{0}$

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The direction of magnetic lines of forces close to a straight conductor carrying current will be :

Along the length of the conductor

Radially outward

Circular in a plane perpendicular to the conductor

Helical

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Explanation

According to the right-hand thumb rule, the direction of magnetic field lines around a straight current-carrying conductor is circular, forming concentric circles perpendicular to the length of the conductor. This pattern results from the circular motion of the charge carriers inside the conductor.

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The magnetic field at the centre of a coil of n turns, bent in the form of a square of side 2 l, carrying current i, is :

$\frac{\sqrt{2} μ_{0} ni}{πl}$

$\frac{\sqrt{2} μ_{0} ni}{2 πl}$

$\frac{\sqrt{2} μ_{0} ni}{4 πl}$

$\frac{2 μ_{0} ni}{πl}$

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Explanation

(a) Magnetic field due to one side of the square at centre O

$B_{1} = \frac{μ_{0}}{4 π} \cdot \frac{2 isin (45 °)}{a / 2} \Rightarrow B_{1} = \frac{μ_{0}}{4 π} \cdot \frac{2 \sqrt{2} i}{a}$

Hence magnetic field at centre due to all side

$B = 4 B_{1} = \frac{μ_{0} (2 \sqrt{2} i)}{πa}$

Magnetic field due to n turns

$B_{net} = nB = \frac{μ_{0} 2 \sqrt{2} ni}{πa} = \frac{μ_{0} 2 \sqrt{2} ni}{π (2 l)} = \frac{\sqrt{2} μ_{0} ni}{πl}$ (âˆµ a=2l)

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In a current carrying long solenoid, the field produced does not depend upon :

Number of turns per unit length

Current flowing

Radius of the solenoid

All of the above

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Explanation

Here $B = μ_{0} n i$

where n is number of turns per unit length $= \frac{N}{l}$

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A straight wire of diameter 0.5 mm carrying a current of 1 A is replaced by another wire of 1 mm diameter carrying the same current. The strength of the magnetic field far away is :

Twice the earlier value

Half of the earlier value

Quarter of its earlier value

Unchanged

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Explanation

(d) The magnetic field is given by $B = \frac{μ_{0}}{4 π} \frac{2 i}{r}$ . It is independent of the radius of the wire.

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