Question

A current of 10A is flowing in a wire of length $1.5m$. A force of $15N$ acts on it when it is placed in a uniform magnetic field of $2T$. The angle between the magnetic field and the direction of the current is:
(A) ${30^ \circ }$
(B) ${45^ \circ }$
(C) ${60^ \circ }$
(D) ${90^ \circ }$

Answer 1

Hint: To solve this question, we have to use the formula of the force on a current carrying wire when it is placed in a uniform magnetic field. Then we have to substitute the given values into that formula to get the required value of the angle between the magnetic field and the direction of the current.
Formula used: The formula used to solve this question is given by
\[\vec F = I\left( {\vec l \times \vec B} \right)\], here $\vec F$ is the force acting on a wire having a length of $l$ and carrying a current of $I$, when it is placed in a uniform magnetic field of $\vec B$.

Complete step by step solution:
Let the angle between the magnetic field be $\theta $.
We know that the force exerted on a current carrying wire when it is placed in a magnetic field is given by
\[\vec F = I\left( {\vec l \times \vec B} \right)\]
Writing the magnitudes on both the sides, we get
\[\left| {\vec F} \right| = I\left| {\left( {\vec l \times \vec B} \right)} \right|\]
$ \Rightarrow F = IlB\sin \theta $
According to the question, $F = 15N$, $I = 10{\text{A}}$, $l = 1.5m$, and $B = 2T$. Substituting these above, we get
$15 = 10 \times 1.5 \times 2 \times \sin \theta $
$15 = 30\sin \theta $
Dividing both the sides by $30$ we get
$\sin \theta = \dfrac{{15}}{{30}}$
$ \Rightarrow \sin \theta = 0.5$
Taking sine inverse both the sides, we finally get
$\theta = {30^ \circ }$
Thus, the angle between the magnetic field and the direction of the current is equal to ${30^ \circ }$.

Hence, the correct answer is option A.

Note: Although the cross product which appears in the expression for force on a current carrying wire is between the length and the magnetic field, it is defined between the magnetic field and the direction of current. This is done because we cannot take the cross product of the current as it is a scalar quantity.