Skip to content
NCERT Exemplar · Q7

Q.A long solenoid has 1000 turns per metre and carries a current of 1 A. It has a soft iron core of μr=1000\mu_r = 1000. The core is heated beyond the Curie temperature, TcT_c.

(a) The H field in the solenoid is (nearly) unchanged but the B field decreases drastically.
(b) The H and B fields in the solenoid are nearly unchanged.
(c) The magnetisation in the core reverses direction.
(d) The magnetisation in the core diminishes by a factor of about 10810^{8}.
Uttarakhand UbseMCQ· 1mImportance★★★★★
56% · 19/34 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Inside the solenoid, H=nIH=nI is fixed entirely by the free (coil) current and does not care about the core, so HH stays essentially unchanged as the core is heated through TcT_c. But B=μ0μrHB=\mu_0\mu_r H collapses drastically because μr\mu_r falls from 10001000 to nearly 11 once the core stops being ferromagnetic - matching options (a) and (d).

Setting up the numbers

Given: n=1000 turns/mn=1000\ \text{turns/m}, I=1 AI=1\ \text{A}, and (below TcT_c) μr=1000\mu_r=1000.

The magnetising field HH

Ampere's law for a long solenoid gives the magnetising field purely from the free current in the coil:

H=nI=(1000)(1)=1000 A/m.H = nI = (1000)(1) = 1000\ \text{A/m}.

This expression never involves the core at all - HH is set only by the current and the winding density, so it is (nearly) unchanged whether the core is ferromagnetic or not. This confirms the first half of option (a).

The magnetic field BB

Below TcT_c (ferromagnetic core, μr=1000\mu_r=1000):

Bbelow=μ0μrH=(4π×10−7)(1000)(1000)≈1.26 T.B_{\text{below}} = \mu_0\mu_r H = (4\pi\times10^{-7})(1000)(1000) \approx 1.26\ \text{T}.

Above TcT_c (core no longer ferromagnetic, μr→\mu_r\to nearly 11 - the material becomes ordinary paramagnetic material):

Babove=μ0(1)H=(4π×10−7)(1000)≈1.26×10−3 T.B_{\text{above}} = \mu_0(1) H = (4\pi\times10^{-7})(1000) \approx 1.26\times10^{-3}\ \text{T}.

So BB drops by roughly a factor of 10001000 - a drastic decrease, while HH is unchanged. This is exactly option (a): "the H field is (nearly) unchanged but the B field decreases drastically." Option (b), that both HH and BB stay nearly unchanged, is therefore false (BB changes hugely).

The magnetisation MM

Using B=μ0(H+M)B=\mu_0(H+M), so M=B/μ0−H=(μr−1)HM = B/\mu_0 - H = (\mu_r-1)H:

  • Below TcT_c: Mbelow=(1000−1)(1000)≈1×106 A/mM_{\text{below}} = (1000-1)(1000) \approx 1\times10^6\ \text{A/m}. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.