Question

A long straight wire carrying current of $30\text{A}$ is placed in an external uniform magnetic field of $4\times {10}^{-4}\text{T}$ parallel to the current. Find the magnitude of net magnetic field at point $2\text{cm}$ away from the wire.

• A. $7\times {10}^{-4}\text{T}$
• B. $1\times {10}^{-4}\text{T}$
• C. $5\times {10}^{-4}\text{T}$
• D. $6\times {10}^{-4}\text{T}$

Answer 1

The correct option is C $5\times {10}^{-4}\text{T}$



Magnetic field at point $\text{P}$ due to the current carrying wireis,

$({B_P)}_W=\frac{{\mu }_0}{2\pi }\frac{I}{a}$

$=2\times {10}^{-7}\times \frac{30}{2\times {10}^{-2}}$

$=30\times {10}^{-5}$

$=3\times {10}^{-4}\text{T}\text{(into the paper plane)}$

External magnetic field at point $\text{P}$ is,

$(B_P{)}_{\text{ext}}=4\times {10}^{-4}\text{T}\text{(in upward direction)}$

$\overrightarrow{B_{net}}=(\overrightarrow{B_P}{)}_W+(\overrightarrow{B_P}{)}_{\text{ext}}$

$\left|B_{net}\right|=\sqrt{(B_P{)}_W^2+(B_P{)}_{\text{ext}}^2+2\left(B_P{)}_W\right(B_P{)}_{\text{ext}}\cos \theta }[\because \theta ={90}^{\circ }]$

$\left|B_{net}\right|=\sqrt{(B_P{)}_W^2+(B_P{)}_{\text{ext}}^2}$

$=\sqrt{(3\times {10}^{-4}{)}^2+(4\times {10}^{-4}{)}^2}$

$=5\times {10}^{-4}\text{T}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (c) is the correct answer.