Question

A conductor of length $0.04\text{m}$ is tangentially connected to a circular loop of radius $0.03\text{m}$ perpendicular to its plane. Find the magnetic field induction at the centre of the loop, if $4\text{A}$ current is passed through the conductor as shown in figure.

• A. $2.2\times {10}^{-5}\text{T}$
• B. $1.07\times {10}^{-5}\text{T}$
• C. $2.75\times {10}^{-5}\text{T}$
• D. $3.25\times {10}^{-5}\text{T}$

Answer 1

The correct option is B $1.07\times {10}^{-5}\text{T}$


Due to the symmetry in the two halves of the loops, the current will distribute equally i.e $2\text{A}$.

Thus, magnetic field due to two equal halves will be equal in magnitude & opposite in direction, hence get cancelled.

$\therefore B_{loop}=0$

And the horizontal wire is lying in the plane of loop and passes through the centre as shown in figure, so magnetic field due to this wire at centre will be zero.

$\therefore B_{horizontalwire}=0$

Magnetic field induction at the centre of the loop due to a straight current carrying conductor will be

$B_{wire}=\frac{{\mu }_{0}i}{4\pi d}[\sin {\varphi }_1+\sin {\varphi }_2]$

$[\because {\varphi }_1=0^{\circ },\sin {\varphi }_2=\frac{0.04}{0.05}]$

$B_{wire}=\frac{4\pi \times {10}^{-7}\times 4}{4\pi \times 0.03}[\sin 0^{\circ }+\frac{4}{5}]$

$B_{wire}=\frac{16}{15}\times {10}^{-5}=1.07\times {10}^{-5}\text{T}$

So, net magnetic field at the centre will be

$B_{net}=B_{wire}=1.07\times {10}^{-5}\text{T}$

Hence, option $\left(b\right)$ is correct.