Question

A wire 2.80 m in length carries a current of 8.00 A in a region where a uniform magnetic field has a magnitude of 0.450 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

(a) 60.0°

(b) 90.0°

(c) 120°

Answer 1

Concepts and reason

Use the concept of magnetic force to solve this problem.

Initially, calculate the magnitude of the magnetic force between the magnetic field and the current at an angle of $60.0^\circ$ by using the relation between the magnetic force, magnetic field, current, length of the wire and the angle between the magnetic field and the current.

Later, find the magnitude of the magnetic force between the magnetic field and the current at an angle of $90.0^\circ$ by using the relation between the magnetic force, magnetic field, current, length of the wire and the angle between the magnetic field and the current.

Finally, determine the magnitude of the magnetic force between the magnetic field and the current at an angle of $120.0^\circ$ by using the relation between the magnetic force, magnetic field, current, length of the wire and the angle between the magnetic field and the current.

Fundamentals

The expression for the magnitude of the magnetic force is as follows:

${F_{\rm{B}}} = BIL\sin \theta$

Here, ${F_{\rm{B}}}$ is the magnetic force, B is the magnetic field, I is the current flowing in the wire, L is the length of the wire, and $\theta$ is the angle between the magnetic field and the current.

(a)

The magnitude of the magnetic force between the magnetic field and the current at an angle of $60.0^\circ$ is,

${\left( {{F_{\rm{B}}}} \right)_{60.0^\circ }} = BIL\sin \theta$

Here, ${\left( {{F_B}} \right)_{60.0^\circ }}$ is the magnetic force at an angle of $60.0^\circ$ .

Substitute 0.450 T for B, 8.00 A, 2.80 m for L, and $60.0^\circ$ for $\theta$ .

$\begin{array}{c}\\{\left( {{F_{\rm{B}}}} \right)_{60.0^\circ }} = \left( {0.450\;{\rm{T}}} \right)\left( {8.00\;{\rm{A}}} \right)\left( {2.80\;{\rm{m}}} \right)\sin \left( {60.0^\circ } \right)\\\\ = 8.73\;{\rm{N}}\\\end{array}$

(b)

The magnitude of the magnetic force between the magnetic field and the current at an angle of $90.0^\circ$ is,

${\left( {{F_{\rm{B}}}} \right)_{90.0^\circ }} = BIL\sin \theta$

Here, ${\left( {{F_B}} \right)_{90.0^\circ }}$ is the magnetic force at an angle of $90.0^\circ$ .

Substitute 0.450 T for B, 8.00 A, 2.80 m for L, and $90.0^\circ$ for $\theta$ .

$\begin{array}{c}\\{\left( {{F_{\rm{B}}}} \right)_{90.0^\circ }} = \left( {0.450\;{\rm{T}}} \right)\left( {8.00\;{\rm{A}}} \right)\left( {2.80\;{\rm{m}}} \right)\sin \left( {90.0^\circ } \right)\\\\ = 10.08\;{\rm{N}}\\\end{array}$

(c)

The magnitude of the magnetic force between the magnetic field and the current at an angle of $120.0^\circ$ is,

${\left( {{F_{\rm{B}}}} \right)_{120.0^\circ }} = BIL\sin \theta$

Here, ${\left( {{F_{\rm{B}}}} \right)_{120.0^\circ }}$ is the magnetic force at an angle of $120.0^\circ$ .

Substitute 0.450 T for B, 8.00 A, 2.80 m for L, and $120.0^\circ$ for $\theta$ .

$\begin{array}{c}\\{\left( {{F_{\rm{B}}}} \right)_{120.0^\circ }} = \left( {0.450\;{\rm{T}}} \right)\left( {8.00\;{\rm{A}}} \right)\left( {2.80\;{\rm{m}}} \right)\sin \left( {120.0^\circ } \right)\\\\ = 8.73\;{\rm{N}}\\\end{array}$

Ans: Part a

The magnitude of the magnetic force between the magnetic field and the current at an angle of $60.0^\circ$ is 8.73 N.

Part b

The magnitude of the magnetic force between the magnetic field and the current at an angle of $90.0^\circ$ is 10.08 N.

Part c

The magnitude of the magnetic force between the magnetic field and the current at an angle of $60.0^\circ$ is 8.73 N.