Karnataka CET 2010 Questions (MCQ) on Magnetic Field due to Current Carrying Conductors

Questions on magnetic field due to current carrying conductors are generally interesting. Four questions were included from this section in the Karnataka CET 2010 question paper. Here are those questions with solution:

(1) Two thick wires and two thin wires, all of same material and same length, form a square in three different ways P, Q and R as shown in the figure:

With correct connections shown, the magnetic field due to the current flow, at the centre of the loop will be zero in the case of ……

(a) Q and R only

(b) P only

(c) P and Q only

(d) P and R only

In all cases the current in the upper branch produces a magnetic field acting normally into the plane of the figure where as the current in the lower branch produces a magnetic field acting normally outwards (towards the reader). The two fields are thus in opposition. But the currents in the two branches are equal in the case of P and R since the resistances are equal. Therefore, the magnetic field due to the current flow, at the centre of the loop will be zero in the case of P and R only [Option (d)].

(2) Magnetic field at the centre of a circular coil of radius R due to a current I flowing through it is B. The magnetic field at a point along the axis at distance R from the centre is……

(a) B/2

(b) B/4

(c) B/√8

(d) √8 B

The magnitude of the magnetic field (B_axis) on the axis of a plane circular coil at a distance x from the centre of the coil is given by

B_axis = (μ₀nR²I) /2(R²+x²)^3/2 where R is the radius of the coil and I is the current in the coil.

Therefore, the magnetic field (B₁) at a point along the axis at distance R from the centre is given by

B₁ = (μ₀nR²I) /2(R²+R²)^3/2 = (μ₀nR²I) /(2×2√2 ×R³)

Or, B₁ = (μ₀nI) /(2R×2√2)

But the magnetic field at the centre of the coil is given by

B = (μ₀nI) /2R

Therefore, we have B₁ = B/2√2 = B/√8 as given in option (c).

(3) A current I is flowing through a loop. The direction of the current and the shape of the loop are as shown in the figure. The magnetic field at the centre of the loop is μ₀I/R times….. (MA = R, MB = 2R, ÐDMA = 90º)

(a) 5/16, but out of the plane of the paper

(b) 5/16, but into the plane of the paper

(c) 7/16, but out of the plane of the paper

(d) 7/16, but into the plane of the paper

The magnetic field at the centre of a single turn plane circular coil carrying current I is given by

B = (μ₀I) /2r where r is the radius

The loop given in the question consists of three-fourths (DIA) of a circular loop of radius R, one fourth (BIC) of a circular loop of radius 2R and two straight wires AB and CD. The straight wires do not produce any magnetic field at the common centre M of the circular arcs. The circular arcs DIA and BIC produce magnetic fields acting normally into the plane of the loop so that they add up to produce the resultant field at M.

Therefore, magnetic field at M = [(¾)× μ₀I /2R] + [(¼ )× μ₀I /2(2R)]

Or, magnetic field at M = (μ₀I/R) [(3/8) + (1/16)] = (μ₀I/R)(7/16)

The correct option is (d).

(4) PQ and RS are long parallel conductors separated by certain distance. M is the mid point between them (see the figure). The net magnetic field at M is B. Now, the current 2A is switched off. The field at M now becomes……

(a) 2 B

(b) B

(c) B/2

(d) 3 B

If the magnitude of the magnetic field produced by the conductor RS (which carries current 1 A) is B, the magnitude of the magnetic field produced by the conductor PQ (which carries current 2 A) must be 2B. But these fields are oppositely directed. That’s why the resultant field at M has magnitude B when both conductors carry currents. When the current 2A is switched off, the field at M becomes B.

[If one of the options were – B, you would pick it out as the correct one].