To determine the magnitude of the magnetic force on a current-carrying wire in a magnetic field, we use the formula for the magnetic force on a current element, which is given by:
d=Id×where:
I is the current,
d is the infinitesimal length vector of the wire, and
is the magnetic field vector.
In this problem, the given parameters are:
Length of the wire,
L=2mCurrent,
I=1A Direction of the wire along the x axis, so
d can be expressed as
dx.
The magnetic field
=B0(++)T.
Firstly, since the wire is along the x axis, its length vector is:
=2mSince it carries a current of 1 A , the force on the wire can be given by integrating along its length:
=I‌∫d×=I×Substituting the values, we get:
=1A×2×B0(++)Using the distributive property of the cross product, we have:
=2B0×(++)Since the cross product of any vector with itself is zero,
×=0 :
=2B0(×+×) We know from the right-hand rule for cross products that:
‌×=‌×=−Substituting these into the equation, we have:
=2B0(−)The magnitude of this force is obtained by finding the vector magnitude:
Thus, the magnitude of the magnetic force on the wire is:
Option D:
2√2B0N