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Question: A circular coil of \[200\] turns and radius \[10\;{\text{cm}}\] is placed in an uniform magnetic fie...

A circular coil of 200200 turns and radius 10  cm10\;{\text{cm}} is placed in an uniform magnetic field of 0.1  T0.1\;{\text{T}} normal to the plane of the coil. The coil carries a current of 5  A5\;{\text{A}}. The coil is made up of copper wire of cross-sectional area 10−5  m2{10^{ - 5}}\;{{\text{m}}^{\text{2}}} and the number of free electrons per unit volume of copper is 1029{10^{29}}. The average force is experienced by an electron in the coil due to magnetic field is
A. 5×10−25  N5 \times {10^{ - 25}}\;{\text{N}}
B. Zero
C. 8×10−24  N8 \times {10^{ - 24}}\;{\text{N}}
D. None of these.

Explanation

Solution

Charge on the electron is given by,
e=1.6×10−19  Ce = 1.6 \times {10^{ - 19}}\;{\text{C}}. The equation for magnetic force is given by,
F=BevdF = Be{v_d}. The equation for drift velocity is given by,
vd=INeA{v_d} = \dfrac{I}{{NeA}}

Complete step by step solution:
Given, Number of turns in the circular coil,
n=20n = 20
Uniform magnetic field strength,
B=0.1  TB = 0.1\;{\text{T}}
Radius of the circular coil,
r=10  cmr = {\text{10}}\;{\text{cm}}
Convert it to the meter unit.
Therefore,
r=0.1  mr = {\text{0}}{\text{.1}}\;{\text{m}}
Current carrying by the coil,
I=5  AI = {\text{5}}\;{\text{A}}
Cross-sectional area of the copper wire,
A=10−5  m2A = {10^{ - 5}}\;{{\text{m}}^{\text{2}}}
Number of free electrons per unit volume of copper,
N=1029N = {10^{29}} Per meter cube.

According to the question, the magnetic field is uniform. Hence, the total torque in the coil is zero.

We know that the amount of charge applied to an object reflects the amount of imbalance on that object between electrons and protons.
Charge on the electron is given by,
e=1.6×10−19  Ce = 1.6 \times {10^{ - 19}}\;{\text{C}}

The magnetic force is a result of the electromagnetic force, and one nature's fundamental forces, and it is caused by the charging motion.
The equation for magnetic force is given by,
F=BevdF = Be{v_d} …… (i)
Where, vd{v_d} is drift velocity of electrons and the formula for drift velocity is given by,
vd=INeA{v_d} = \dfrac{I}{{NeA}}
Now replace the value of vd{v_d} in equation (i)
Therefore,

F = \dfrac{{BeI}}{{NeA}} \\\ F = \dfrac{{BI}}{{NA}} \\\ $$ …… (ii) Now substitute the values of $$B$$, $$I$$, $$N$$, and $$A$$ in equation (ii)

F = \dfrac{{0.1;{\text{T}} \times {\text{5}};{\text{A}}}}{{{{10}^{29}} \times {{10}^{ - 5}};{{\text{m}}^{\text{2}}}}} \\
= 5 \times {10^{ - 25}};{\text{N}} \\

**Hence, the average force on each electron is $$5 \times {10^{ - 25}}\;{\text{N}}$$. The correct option is A.** **Note:** In order to calculate the average force we use the formula for magnetic force which is given by, $$F = Be{v_d}$$. Where, $${v_d}$$ is drift velocity of electrons and the formula for drift velocity is given by, $${v_d} = \dfrac{I}{{NeA}}$$.