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Question: A solenoid of length 0.5 m and diameter 0.6 cm consists of 1000 turns of fine wire carrying a curren...

A solenoid of length 0.5 m and diameter 0.6 cm consists of 1000 turns of fine wire carrying a current of 5.0×10−35.0 \times {10^{ - 3}} ampere. The magnetic field in Weber/ m2{m^2} at the ends of the solenoid will be-
A. 7.8×10−67.8 \times {10^{ - 6}}
B. 6.28×10−66.28 \times {10^{ - 6}}
C. 3.14×10−63.14 \times {10^{ - 6}}
D. 6.28×10−56.28 \times {10^{ - 5}}

Explanation

Solution

Magnetic field will be equal on both ends of the solenoid. So the magnetic field at the end of a long solenoid can be calculated by using the formula below, where B is the magnetic field, I is the current, N is the ratio of the number of turns over its length. Use this info to find the magnetic field.

Complete step by step answer:
Magnetic field at the end of a solenoid is B=μ0NI2B = \dfrac{{{\mu _0}NI}}{2}
We are given that a solenoid of length 0.5 m and diameter 0.6 cm consists of 1000 turns of fine wire carrying a current of 5.0×10−35.0 \times {10^{ - 3}} ampere.
We have to calculate the magnetic field in Weber/ m2{m^2} at the ends of the solenoid.

Magnetic field is a vector quantity which describes the magnetic effect on moving electric charges and magnetic materials. A charge which is moving in a magnetic field experiences a force perpendicular to the magnetic field and to its velocity.
Magnetic field can be found using B=μ0NI2B = \dfrac{{{\mu _0}NI}}{2}
Where N is the ratio of no. of turns in the solenoid to the length of the solenoid
No. of turns is 1000 and the length is 0.4 m
N=nL ⇒N=10000.4 ⇒N=2500  N = \dfrac{n}{L} \\\ \Rightarrow N = \dfrac{{1000}}{{0.4}} \\\ \Rightarrow N = 2500 \\\
μ0{\mu _0} is the magnetic constant or the permeability of free space and its value is 4π×10−7H/m4\pi \times {10^{ - 7}}H/m
On substituting all the given and obtained values in the magnetic field formula, we get
B=4×π×10−7×2500×5×10−32 →B=157079.63×10−102=78539.81×10−10 ∴B=7.8×10−6Weber/m2  B = \dfrac{{4 \times \pi \times {{10}^{ - 7}} \times 2500 \times 5 \times {{10}^{ - 3}}}}{2} \\\ \to B = \dfrac{{157079.63 \times {{10}^{ - 10}}}}{2} = 78539.81 \times {10^{ - 10}} \\\ \therefore B = 7.8 \times {10^{ - 6}}Weber/{m^2} \\\
Therefore, the correct option is Option A, 7.8×10−67.8 \times {10^{ - 6}}

Note: Do not confuse the magnetic field of the solenoid at its centre with the magnetic field of the solenoid at its ends. The magnetic field at the centre of a solenoid is twice the magnetic field at the ends. One weber unit is the product of one Henry unit with one ampere unit.