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Question: A semicircular arc of radius r and a straight wire along the diameter, both are carrying the same cu...

A semicircular arc of radius r and a straight wire along the diameter, both are carrying the same current i. Find out magnetic force per unit length on the small element P, which is at the centre of curvature?

A. (μ0i24r)\left( \dfrac{{{\mu }_{0}}{{i}^{2}}}{4r} \right)
B. (μ0i22r)\left( \dfrac{{{\mu }_{0}}{{i}^{2}}}{2r} \right)
C. (μ0i2r)\left( \dfrac{{{\mu }_{0}}{{i}^{2}}}{r} \right)
D. (2μ0i2r)\left( \dfrac{2{{\mu }_{0}}{{i}^{2}}}{r} \right)

Explanation

Solution

Hint: Biot-Savart’s law confirms that there won’t be any magnetic field at P due to the straight wire. Because, the point lies on the wire. We need to find the magnetic field at P due to the semicircular arc. We can find the force using the formula,
F=BilF=Bil

Where,
BB is the magnetic field at that location
ii is the current
ll is the length of the wire.

Formula Used:
Magnetic field due to the semicircular arc,
B=μ0i4rB=\dfrac{{{\mu }_{0}}i}{4r}

Where,
μ0{{\mu }_{0}} is the permittivity of the material
ii is the current through the arc
rr is the radius of the arc

Force per unit length is,
Fl=Bi\dfrac{F}{l}=Bi

Where,
BB is the magnetic field
ii is the current through the point

Complete step by step answer:
The radius of the semicircular arc is = r

Current passing through the loop = i

We can find the magnetic field due to a semicircular arc at the centre using Biot-Savart’s law. It is given by,
B=μ0i4rB=\dfrac{{{\mu }_{0}}i}{4r}

Where,
μ0{{\mu }_{0}} is the permittivity of the material
ii is the current through the arc
rr is the radius of the arc

So, at point P the magnetic field is,
B=μ0i4rB=\dfrac{{{\mu }_{0}}i}{4r}

Now, the current flowing through wire at P is i

The force per unit length due to the Magnetic field on a wire is,
Fl=Bi\dfrac{F}{l}=Bi...................(1)

Where,
FF is the force on the wire
BB is the magnetic field at that location
ii is the current going through the wire
ll is the length of the wire

So, in this case, the force per unit area will be given by,
Fl=Bi\dfrac{F}{l}=Bi
Fl=(μ0i4r)i\Rightarrow \dfrac{F}{l}=(\dfrac{{{\mu }_{0}}i}{4r})i
Fl=μ0i24r\Rightarrow \dfrac{F}{l}=\dfrac{{{\mu }_{0}}{{i}^{2}}}{4r}

So, the magnetic force per unit length is
μ0i24r\dfrac{{{\mu }_{0}}{{i}^{2}}}{4r}

Hence, the correct option is (A).

Note: You can divide the entire loop into different segments. Find the magnetic field due to each segment. Don’t consider the segments which are aligned with the point at which we are finding the magnetic field. After that we can use the simple equation (1) to find force per unit length.