Question

A wire as shown in the figure carries a current of 2A.The semicircle has radius of 4cm, the magnetic field at the center C will be

• A. $\frac{3\pi }{2}\times {10}^{-7}T$
• B. $\frac{4\pi }{3}\times {10}^{-7}T$
• C. $\frac{\pi }{2}\times {10}^{-5}T$
• D. $\frac{6\pi }{5}\times {10}^{-5}T$

Answer 1

The correct option is C $\frac{\pi }{2}\times {10}^{-5}T$

Since the point C is in line with the straight parts of the wire, the magnetic field due to the straight parts of the wire will be zero. Therefore, the magnetic field is only due to the circular part.
The magnetic field at the centre of a circular loop is given as
$B=\frac{{\mu }_{0}I}{2r}Here,B=\frac{4\pi \times {10}^{-7}\times 2}{2\times 4\times {10}^{-2}B=\pi \times {10}^{-5}}T\text{Here only half of the circuilar loop}\text{is present so the magnetic field at C}\text{will be:}{B}_c=\frac{\pi }{2}\times {10}^{-5}T$