Question

A 13.0g wire of length L = 62.0 cm is suspended by a pair of flexible leads in a uniform magnetic field of magnitude 0.440T (Fig. 28-41). What are the (a) magnitude and (b) direction (left or right) of the current required to remove the tension in the supporting leads?

Short answer

1. Magnitude of the current is 0.467 A
2. Direction of the current is from left to right

Step-by-step solution

Listing the given quantities

Mass of the wire, $m=13.0g\left(\frac{{10}^{-3}kg}{1g}\right)=0.013kg$

Magnetic field, B = 0.440 T

Length of the rod,$L=62.0cm\left(\frac{{10}^{-3}m}{1cm}\right)=0.62m$

to understand the concept

The problem is deals with the calculation of magnitude and direction of the current using right-hand rule. This is a convenience method for quickly finding the direction of a cross-product of 2 vector here the magnetic force on the wire must be in the upward direction, and it must be balanced by the gravitational force of the rod. So, by equating the two forces, we can find the magnitude of the current. To find the direction of the current, we have to use the right-hand rule.

Formula:

Magnetic force, $F_B=iLB\sin \theta$

Magnetic force in vector form $\stackrel{\rightharpoonup }{F_B}=i\stackrel{\rightharpoonup }{L}\times \stackrel{\rightharpoonup }{B}$

Gravitational force F = mg

(a) To calculate magnitude of current

The magnetic force is given by

$F_B=iLB\sin \theta$

Where

$i=Current,L=Lengthoftheconductor,B=Magneticfield,\theta =Anglebetweencurrentandfield$

And the gravitational force is

F = mg

Since both the forces must be balanced, so we can write,

$mg=iLB\sin \theta$

Since the magnetic field and the current are perpendicular to each other, therefore,

$\begin{array}{rcl}\sin \theta & =& \sin {90}^o\\ & =& 1\end{array}$

Thus, the current is

$$\begin{gathered} i=\frac{mg}{LB} \\ i=\frac{\left(0.0130kg\right)\left(9.8m/s^2\right)}{\left(0.620m\right)\left(0.440T\right)} \\ i=0.467A \end{gathered}$$

(b) To find the direction of the current

Magnetic force in vector form is given as

$\stackrel{\rightharpoonup }{F_B}=i\stackrel{\rightharpoonup }{L}\times \stackrel{\rightharpoonup }{B}$

By using the right-hand rule, we can say that the direction of the current is from left to right.