Question

A long straight wire carrying of 30 A is placed in an external uniform magnetic field of induction $4\times {{10}^{-4}}$T. The magnetic field is acting parallel to the direction of current. The magnitude of the resultant magnetic induction in tesla at a point 2.0 cm away from the wire is
A. ${{10}^{-4}}$
B. $3\times {{10}^{-4}}$
C. $5\times {{10}^{-4}}$
D. $6\times {{10}^{-4}}$

Answer 1

Hint:A long straight wire is carrying some current so there will be a magnetic field associated with it. But it is also placed in an external magnetic field. Therefore, the resultant magnetic field at a point of some distance will be the vector sum of both the magnetic fields.



Formula used:

Magnetic field due to a current carrying element is given as:
${{B}_{1}}=\frac{{{\mu }_{0}}i}{2\pi r}$
Where ${{\mu }_{0}}$is the permeability of free space, $i$is the current carried by the wire and r is the distance from the wire to the point.
If both magnetic fields are perpendicular to each other, then magnitude of the resultant magnetic field will be:
$B=\sqrt{B_{1}^{2}+B_{2}^{2}}$


Complete answer:
It is given in this question that a straight long wire is carrying a current of 30A. It is placed in an external magnetic field and we know that due to the current flowing through the wire it possesses a magnetic field. So, at a point which is 2 cm away experiences a resultant of both the fields.
From Ampere’s law, magnetic field around a current carrying long straight wire is:
${{B}_{1}}=\frac{{{\mu }_{0}}i}{2\pi r}=\frac{4\pi \times {{10}^{-7}}\times 30}{2\pi \times 2\times {{10}^{-2}}}=3\times {{10}^{-4}}T$ which is perpendicular to the plane containing current.
External magnetic field is already given in the question, that is:
${{B}_{2}}=4\times {{10}^{-4}}T$
Both fields are perpendicular to each other. Therefore, the magnitude of the resultant magnetic field is:
$B=\sqrt{B_{1}^{2}+B_{2}^{2}}$
On substituting values of magnetic field in the above equation we get:
$B=\sqrt{{{(3\times {{10}^{-4}})}^{2}}+{{(4\times {{10}^{-4}})}^{2}}}=5\times {{10}^{-4}}T$
Therefore, the answer is option (C)




Note:We are using Ampere’s law to find the magnitude of the magnetic field around a long straight wire carrying current. Remember the right hand thumb rule is used to find the direction of the magnetic field. It is given in question that the external magnetic field is parallel to the direction of current. Therefore, both magnetic fields are perpendicular to each other.