Skip to content
PhysicsNEET 2024Biot–Savart and Ampère's laws

Moving Charges and Magnetism

A tightly wound 100-turn coil of radius 10 cm carries a current of 7 A. The magnitude of the magnetic field at the centre of the coil is (take the permeability of free space as 4π×10−74\pi \times 10^{-7} SI units):

  1. A.4.4 T
  2. B.4.4 mT✓
  3. C.44 T
  4. D.44 mT

Solution

B=μ0NI2r=4π×10−7×100×72×0.1=4.4×10−3B = \frac{\mu_0 N I}{2r} = \frac{4\pi\times10^{-7} \times 100 \times 7}{2 \times 0.1} = 4.4 \times 10^{-3} T =4.4= 4.4 mT.

Save for spaced revision (free account)

More from this chapter in past papers

From Physics › Moving Charges and Magnetism › Biot–Savart and Ampère's laws — lectures, notes and every past question on it.

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

ShareWhatsAppTelegram