A circular coil of radius R having N number of turns carries a steady current I. The magnetic induction at the centre of the coil is 0·1 tesla. If the number of turns is doubled and the radius is halved, which one of the following will be the correct value for the magnetic induction at the centre of the coil?
- (a)0·05 tesla
- (b)0·2 tesla
- (c)0·4 tesla
- (d)0·8 tesla
Correct — C, 0·4 tesla. The magnetic field at the centre of a circular coil of N turns and radius R carrying current I is B = μ₀NI / 2R. The current is unchanged, so only the two factors that were altered matter. Doubling the number of turns doubles the numerator, and halving the radius halves the denominator, which doubles the field again. The two changes multiply, giving a field four times the original: 4 × 0·1 = 0·4 tesla. Written out, the new field is μ₀(2N)I / (2 × R/2) = 2μ₀NI / R = 4 × (μ₀NI / 2R).
- (a)0·05 tesla — This halves the original field. It is what you get by treating the smaller radius as though it weakened the field, when in fact B varies inversely with R, so a smaller coil concentrates the field and strengthens it.
- (b)0·2 tesla — This applies only one of the two changes — the doubling of the turns — and forgets that halving the radius doubles the field a second time.
- (d)0·8 tesla — This is eight times the original, which would follow if the field went as the inverse square of the radius. At the centre of a single circular loop the dependence is inverse first power, so the radius change contributes a factor of two, not four.
A current-carrying wire sets up a magnetic field around it, and bending the wire into a loop gathers that field into a strong, roughly uniform region at the centre. Applying the Biot–Savart law to a full circular loop gives B = μ₀I / 2R at the centre, and stacking N identical turns simply multiplies the result by N. The direction is given by the right-hand thumb rule: curl the fingers along the current and the thumb points along the field through the loop.
Numerical questions of this kind never need the value of μ₀ or of the current, because everything except the changed quantities cancels in the ratio. Write the formula, mark which symbols move and which stay, and multiply the factors. Here N goes up by two and R goes down by two, and since R sits in the denominator both changes push the field the same way — hence four, not one. The commonest slips are to apply only one change, or to assume an inverse-square law because so many other fields obey one. Note that the coil's field at the centre falls off as 1/R, while the field far away on the axis of the same loop falls off as 1/x³.
- For a circular coil of N turns, the field at the centre is B = μ₀NI / 2R.
- The field is directly proportional to both the number of turns and the current, and inversely proportional to the radius.
- Doubling N and halving R together multiply the field by four, giving 0·4 tesla from 0·1 tesla.
- The tesla is the SI unit of magnetic flux density; the direction of the field follows from the right-hand thumb rule.
Both alterations push the field the same way, so the factors multiply to four.
- Applying only one of the two stated changes and stopping at double the original value.
- Assuming an inverse-square dependence on radius; at the centre of a loop the dependence is inverse first power.
- Reaching for the value of μ₀ or of the current, neither of which is needed once the problem is set up as a ratio.
Asked as a proportional-change numerical where two quantities are altered at once and the factors have to be combined.
In a solenoid, the current flowing through the wire is I and number of turns per unit length is n. This gives a magnetic field B inside the solenoid. If number of turn per unit length is increased to 2n, what will be the value of magnetic field in the solenoid ?
- (a) B
- (b) 2B
- (c) B/2
- (d) B/4
Answer(b) 2B
The same proportional reasoning for the other standard current geometry. There only the turn density changes, so the field doubles; here two quantities change and the factors multiply.
Which one of the following statements regarding a current-carrying solenoid is not correct?
- (a) The magnetic field inside the solenoid is uniform.
- (b) The current-carrying solenoid behaves like a bar magnet.
- (c) The magnetic field inside the solenoid increases with increase in current.
- (d) If a soft iron bar is inserted inside the solenoid, the magnetic field remains the same.
Answer(d) If a soft iron bar is inserted inside the solenoid, the magnetic field remains the same.
Collects the qualitative dependences that lie behind the formula — field rises with current and with turns, and inserting a ferromagnetic core raises it further.
- practice — not a real PYQ
The magnetic field at the centre of a circular coil is B. If the current is doubled and the radius is also doubled, the new field will be
- (a)B/2
- (b)B
- (c)2B
- (d)4B
Answer(b) B — doubling the current doubles the field and doubling the radius halves it, so the two changes cancel exactly.
- practice — not a real PYQ
The magnetic field at the centre of a circular coil carrying a steady current is
- (a)directly proportional to the radius of the coil
- (b)inversely proportional to the radius of the coil
- (c)independent of the radius of the coil
- (d)inversely proportional to the square of the radius
Answer(b) inversely proportional to the radius of the coil — from B = μ₀NI/2R, so a tighter coil gives a stronger field at its centre.