Question

A. A wire along the x-axis, of length 1, carries current I in the negative x-direction. The magnetic field in the region is nonuniform, it depends on is x such that В: Bo7 for 0 5xs1 otherwise = 0 Here, Bo, is a constant. i. Sketch a graph for B in the interval –2i < x < 21. ji. What is the force exerted on the wire? iii. What is net torque exerted on the wire about the point x = 0?

Answer 1

Refer to the diagram:

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$B = \frac{B_{o}x}{l} \text{ for } 0<x<l$ towards +ve z-axis.

= 0 otherwise

l = length of wire

i = current flowing through wire towards -ve x-axis.

a) The magnetic field is only between x = 0 to x = l and is zero elsewhere.

See the diagram:

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The graph is a straight line passing through origin and slope is $\frac{B_{o}}{l}$ .

b) Magnetic force on a current-carrying wire is given by $F=iLB$ [where L = length of wire perpendicular to the uniform magnetic field B]

Here, we have a non-uniform magnetic field, but for a very small section of wire, we can the magnetic field is uniform.

Consider a small element at a distance 'x' from origin and length 'dx'.

force on the element is -----(i)

Therefore, on integration, we get the net force on the wire,

$F = \int i(dx)B = \int_{0}^{l} i\frac{B_{o}x}{l}(dx)=\frac{iB_{o}}{l}\int_{0}^{l}x*dx = \frac{iB_{o}l}{2}$ [answer]

The direction of force is determined from the vector equation of force $\vec{F}=i(\vec{L}\times \vec{B})$

The direction of this net force is along +ve y-axis.

c) The torque on the element due to the force from equation (i):

$d\tau = x*dF$

$\Rightarrow d\tau = x*iB*dx = x*i\frac{B_{o}x}{l}*dx = \frac{iB_{o}}{l}*x^{2}dx$

Therefore, on integration, we get the net torque on the wire,

$\tau = \frac{iB_{o}}{l}*\int_{0}^{l}x^{2}dx=\frac{iB_{o}}{l}*\frac{l^{3}}{3} = \frac{iB_{o}l^{2}}{3}$ [answer]