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Question: There are 50 turns of a wire in every cm length of a long solenoid. If 4 ampere current is flowing i...

There are 50 turns of a wire in every cm length of a long solenoid. If 4 ampere current is flowing in the solenoid, the approximate value of magnetic field along its axis at an internal point and at one end will be respectively

A

12.6×10−3Weber/m2,6.3×10−3Weber/m212.6 \times 10 ^ { - 3 } \mathrm { Weber } / \mathrm { m } ^ { 2 } , 6.3 \times 10 ^ { - 3 } \mathrm { Weber } / \mathrm { m } ^ { 2 }

B

C

D

Answer

Explanation

Solution

The magnetic field in the solenoid along its axis (i) At an internal point=μoni= \mu _ { o } n i

=4π×10−7×5000×4=25.1×10−3 Wb/m2= 4 \pi \times 10 ^ { - 7 } \times 5000 \times 4 = 25.1 \times 10 ^ { - 3 } \mathrm {~Wb} / \mathrm { m } ^ { 2 }

(Here n=50n = 50 turns /cm=5000/ \mathrm { cm } = 5000 turns /m/ \mathrm { m } )

(ii) At one end

=12.6×10−3 Wb/m2= 12.6 \times 10 ^ { - 3 } \mathrm {~Wb} / \mathrm { m } ^ { 2 }