A solenoid of n turns carries current I while another solenoid of same length and of 3 n turns carries current 4I. What is the ratio of the magnetic strengths of the two solenoid ?
12times
10times
7times
none of the above
A
7times
B
10times
C
none of the above
D
12times
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Solution
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The magnetic field inside a current carrying solenoid is given by
B=μ0nI
where n is the number of turns per unit lenght in the solenoid
I is the current through the wire
Hence B′B=n′I′nI=3×4=12
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