
Provided below are some handwritten notes on magnetic field in a solenoid and other important concepts. The current flowing through the coil can be expressed by

If it has 300 turns, determine the current flowing through it. Faradays law: negative N times our change in flux, and were talking about our change in flux over change in, let me write that a little bit neater, our change in flux over change in time. Im just gonna rewrite Faradays law right over here. The magnetic field generated by the solenoid is 8.505 × 10 −4 N/Amps m.Įxample 2: A solenoid of diameter 40 cm has a magnetic field of 2.9 × 10 −5 N/Amps m. Lets think about what that electromotive force thats going to be induced is going to be. The magnetic field in a solenoid formula is given by, Ampere’s Circuital Law Video ExplanationĪlso Read: Moving Charges and Magnetism MCQĮxample 1: Find out the magnetic field produced by the solenoid of length 80 cm under the number of turns of the coil is 360 and the current passing through is 15 A. The above equation is used to determine the magnitude of the magnetic field in a long solenoid. “The integration of a magnetic field around a loop is directly proportional to the current enclosed in the loop.” Since the field outside the solenoid is comparatively less than that it is present inside, we can consider it as zero as the length of the solenoid increases, and hence B = 0.Īlso Read: Parallel Current Carrying Conductor The magnetic field outside a solenoid is: For the entire number of field lines to be contained, the magnetic field outside must be zero, because the solenoid gets longer.As a result of which, the field outside the solenoid is constant.

