Courses
Courses for Kids
Free study material
Offline Centres
More
Store

The magnetic force on a current carrying wire placed in uniform magnetic field if the wire is oriented perpendicular to magnetic field is:(A) $0$(B) $ILB$(C) $2ILB$(D) $ILB/2$

Last updated date: 20th Jun 2024
Total views: 54.3k
Views today: 1.54k
Verified
54.3k+ views
Hint: The magnetic field due to a flowing charge or current carrying conductor is the product of the magnetic field at that point, the current flowing through the conductor and the length of the wire. It depends on the sine of the angle made by the magnetic field and the length of the wire.

Complete step by step solution:
In a magnetic field a current carrying wire experiences a magnetic force, which is given by –
$\vec F = I(d\vec l \times \vec B)$
Where I is the magnitude of current flowing in the conductor,
$dl$is the length of the wire
And B is the magnetic field.
Since the Force is also a vector quantity, the product between the length of wire and magnetic field is a cross product. Therefore, this formula can be rewritten as-
$\vec F = IBL\operatorname{Sin} \theta$
Where$\theta$is the angle between the magnetic field and the orientation of the wire in that field.
In the question we have,
$\theta = 90^\circ$as the wire is perpendicular to the magnetic field.
Therefore the force is given by,
$\vec F = IBL\operatorname{Sin} 90^\circ$
Since, $\sin 90^\circ = 1$
$F = IBL \times 1$
$F = IBL$

Thus, the answer is option (B).

$\vec F = q\left( {\vec v \times \vec B} \right)$
$q\vec v = dq\dfrac{{dl}}{{dt}} = \dfrac{{dq}}{{dt}}dl = IL$