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Question: What is the net force on the square coil: A) \(25 \times 10^{-7} N\) moving towards the wire B) ...

What is the net force on the square coil:
A) 25×10−7N25 \times 10^{-7} N moving towards the wire
B) 25×10−7N25 \times 10^{-7}N moving away from the wire
C) 35×10−7N35 \times 10^{-7} N moving towards the wire
D) 35×10−7N35 \times 10^{-7}N moving away from the wire

Explanation

Solution

When an electric current passes through a wire, it creates a magnetic field. This magnetic field exerts a magnetic force on a current-carrying wire in a direction given by the right-hand rule. Also, the direction of the magnetic field depends upon the direction of the current. And the magnitude of the force depends on the current in the conductor, the strength of the magnetic field, and the length of the conductor in the magnetic field.

Complete step by step solution:
From the above figure, it is clear that force on side BC and force on side AD are equal but opposite in direction. Hence their net force will be zero.
Now let us calculate the force on side AB.
We know that the force on a wire is given by the formula, F == B I L where
B == magnetic field in Tesla
I == Current in Ampere
L == Length in meter
∴\therefore Force on side AB, FAB=BILF_{AB} = B I L
We also know that the magnetic field is given by the formula, BB = μ0I2πr\dfrac{μ_0 I}{2 \pi r}
Where μ0μ_0 == Permeability of free space (4π×10−7T.m/A)(4 \pi \times 10^{-7} T. m / A)
rr = distance in meters
Thus, substituting the values of B, I, and L for force on side AB, we get,
FABF_{AB} = 4×π×10−7×2×0.152×π×0.02\dfrac{{4 \times \pi \times {{10}^{ - 7}} \times 2 \times 0.15}}{{2 \times \pi \times 0.02}} (I=2A,L=15cm=0.15m(I = 2 A, L = 15 cm = 0.15 m and r=2cm=0.02m)r = 2 cm = 0.02 m)
⇒\Rightarrow FAB=30×10−7NF_{AB} = 30 \times 10^{-7} N
∴\therefore FAB=3×10−6NFAB = 3 \times 10^{-6}N
Similarly force on side CD, FCDF_{CD} == B I L
So, substituting the values of B, I, and L for force on side CD, we get,
FCDF_{CD} == 4×π×10−7×2×0.152×π×0.12\dfrac{{4 \times \pi \times {{10}^{ - 7}} \times 2 \times 0.15}}{{2 \times \pi \times 0.12}}
⇒\Rightarrow FCD=5×10−7NF_{CD} = 5 \times 10^{-7} N
∴\therefore FCD=0.5×10−6NF_{CD} = 0.5 \times 10^{-6} N
∴\therefore Net force, FnetF_{net} = FAB−FCDF_{AB} - F_{CD}
⇒\Rightarrow FnetF_{net} = 3×10−6N−0.5×10−6N3 \times 10^{-6}N - 0.5 \times 10^{-6} N
⇒\Rightarrow FnetF_{net} = 2.5×10−6N 2.5 \times 10^{-6}N
∴\therefore FnetF_{net} = 25×10−7N25 \times 10^{-7}N moving towards the wire
∴\therefore Net force, FnetF_{net} = 25×10−7N25 \times 10^{-7}N moving towards the wire

Hence the correct option for the problem is A.

Note: 1) μ0μ_0, Permeability of free space is constant and its value is 4π\pi ×10−7T.m/A. \times 10^{-7} T. m / A.
2) Direction of force is reversed when the direction of current through the conductor is reversed.
3) If the direction of the magnetic field is reversed, the direction of force gets reversed.