EXAMPLE 3.18
Compute the magnitude of the magnetic field of a long, straight wire carrying a current of 1A at distance of 1m from it. Compare it with Earth’s magnetic field.
Solution
Given that 1 = 1 A and radius r = 1 m
But the Earth’s magnetic field is BEarth ≈ 10−5 T
So, Bstraightwire is one hundred times smaller than BEarth.
EXAMPLE 3.19
Calculate the magnetic field inside a solenoid, when
(a) the length of the solenoid becomes twice and fixed number of turns
(b) both the length of the solenoid and number of turns are double
(c) the number of turns becomes twice for the fixed length of the solenoid
Compare the results.
Solution
The magnetic field of a solenoid (inside) is
(a) length of the solenoid becomes twice and fixed number of turns
L→2L (length becomes twice)
N→N (number of turns are fixed)
The magnetic field is
B2L , N = µ NI/2L = 1/2 BL ,N
(b) both the length of the solenoid and number of turns are double
L→2L (length becomes twice)
N→2N (number of turns becomes twice)
The magnetic field is
(c) the number of turns becomes twice but for the fixed length of the solenoid
L→L (length is fixed)
N→2N (number of turns becomes twice)
The magnetic field is
BL ,2 N = µ, 2NI/L = 2BL ,N
From the above results,
BL ,2 N > B2 L ,2 N > B2 L , N
Thus, strength of the magnetic field is increased when we pack more loops into the same length for a given current.
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