Class 12 Physics
CBSE and UP Board
Moving Charges and Magnetism
Chapter-4 Exercise 4.15
Additional Exercise
NCERT Solutions Class 12 Physics Chapter 4 Question 15
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Magnetic field strength, B = 100 G = 100 x 10⁻⁴ T
Number of turns per unit length, n = 1000 turns m⁻1
Current flowing in the coil, I = 15 A
Permeability of free space, μ0 = 4π x 10-7 T m A-1
Magnetic field is given by the relation,
B = μ0 nl
=>nl =B/μ0 = (100 x 10⁻⁴ )/(4π x 10-7 )
= 7957.74 ≈ 8000 Am⁻¹
If the length of the coil is taken as 50 cm, radius 4 cm, number of turns 400, and current 10 A, then these values are not unique for the given purpose. There is always a possibility of some adjustments with limits.