CBSE Class 12-science Answered
Apply Ampere circuital law to find the magnetic field inside and outside of a solenoid
Asked by ap996969 | 23 Feb, 2019, 09:21: AM
Expert Answer
Figure shown above represents this idealised picture. The field outside the solenoid approaches zero.
We shall assume that the field outside is zero. The field inside becomes everywhere parallel to the axis.
Consider a rectangular Amperian loop abcd. Along cd the field is zero. Along transverse sections bc and ad, the field component
is zero. Thus, these two sections make no contribution.
is zero. Thus, these two sections make no contribution.
Let the field along ab be B. Thus, the relevant length of the Amperian loop is, L = h.
Let n be the number of turns per unit length, then the total number of turns is n×h.
The enclosed current is, Ie = I (n × h), where I is the current in the solenoid. From Ampere’s circuital law
B×L = μ0 × Ie, B × h = μ0I(n × h)
B = μ0 n I
The direction of the field is given by the right-hand rule.
B×L = μ0 × Ie, B × h = μ0I(n × h)
B = μ0 n I
The direction of the field is given by the right-hand rule.
Answered by Thiyagarajan K | 23 Feb, 2019, 01:56: PM
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