CBSE Class 12 · 2008 · All India · Magnetic Effects of CurrentShort Answer

Using Ampere's circuital law, derive an expression for the magnetic field along the axis of a toroidal solenoid.

Answer: $B = μ_{0} n I$ , where $n$ is the number of turns per unit length.

Solution

Take a circular amperian loop of radius

r

along the axis of the toroid, concentric with it. By symmetry

B

is tangential and constant in magnitude along this loop, so

\oint B \cdot d ℓ = B (2 π r) .

If the toroid has

N

turns carrying current

I

, the current threading the loop is

N I

, so by Ampere's law

B (2 π r) = μ_{0} N I \Rightarrow B = \frac{μ _{0} N I}{2 π r} = μ_{0} n I,

writing

n = \frac{N}{2 π r}

for the number of turns per unit length. The field is confined to the core; it is zero outside.

CBSE Class 12 Board · 2008 · All India · Subjective