A solenoid of n turns carries current I while another solenoid of same length and of $3$ n turns carries current $4 I .$ What is the ratio of the magnetic strengths of the two solenoid ?
$12 t i m e s$
$10 t i m e s$
$7 t i m e s$
none of the above

Question

A solenoid of n turns carries current I while another solenoid of same length and of $3$ n turns carries current $4 I .$ What is the ratio of the magnetic strengths of the two solenoid ?
$12 t i m e s$
$10 t i m e s$
$7 t i m e s$
none of the above

Toppr Admin · Accepted Answer

The magnetic field inside a current carrying solenoid is given by-xA0-B-x3BC-0nIwhere n is the number of turns per unit lenght in the solenoidI is the current through the wireHence B-x2032-B-n-x2032-I-x2032-nI-3-xD7-4-12

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 ?12times10times7timesnone of the above

The magnetic field inside a current carrying solenoid is given by B=μ0nIwhere n is the number of turns per unit lenght in the solenoidI is the current through the wireHence B′B=n′I′nI=3×4=12

A solenoid of n turns carries current I while another solenoid of same length and of $3$ n turns carries current $4 I .$ What is the ratio of the magnetic strengths of the two solenoid ?
$12 t i m e s$
$10 t i m e s$
$7 t i m e s$
none of the above

The magnetic field inside a current carrying solenoid is given by
$B = μ_{0} n I$
where $n$ is the number of turns per unit lenght in the solenoid
$I$ is the current through the wire
Hence $\frac{B^{'}}{B} = \frac{n^{'} I^{'}}{n I} = 3 \times 4 = 12$