2020 Moving Charges and MagnetismMagnetic Field of a Solenoid Easy Question 2 of 3 in Magnetic Field of a Solenoid

A long solenoid of 50 cm length having 100 turns carries a current of 2.5 A. The magnetic field at the centre of the solenoid is: ($\mu_0=4\pi\times10^{-7}\ T\,m\,A^{-1}$)

A $3.14\times10^{-4}\ T$
B $6.28\times10^{-5}\ T$
C $3.14\times10^{-5}\ T$
D $6.28\times10^{-4}\ T$

The answer is D.

Solution

$B = \mu_0 nI$, with $n = \frac{N}{L} = \frac{100}{50 \times 10^{-2}} = 200$ turns/m $B = 4\pi \times 10^{-7} \times 200 \times 2.5 = 6.28 \times 10^{-4}$ T

Concept B = μ0 n I

Magnetic Field of a Solenoid in past papers

3 questions from this chapter have appeared across 3 exam years.