. A wire having a linear mass density of 1.00 g / cm is placed on a horizontal surface that has

step 1 Hello friends here it is given. A wire having linear mass density 1 gram per centimeter or 0 .1

. A wire having a linear mass density of 1.00 g / cm is...

. A wire having a linear mass density of $1.00 \mathrm{g} / \mathrm{cm}$ is placed on a horizontal surface that has a coefficient of kinetic friction of $0.200 .$ The wire carries a current of 1.50 A toward the east and slides horizontally to the north. What are the magnitude and direction of the smallest magnetic field that enables the wire to move in this fashion?

Key Concepts

Force Equilibrium

Force equilibrium in physics implies that all the forces acting on a body balance out, resulting in zero net force and typically a constant velocity. In the context of the problem, this concept is used to set the magnetic force equal to the kinetic friction force, identifying the minimum strength and direction of the magnetic field required for steady motion.

Kinetic Friction

Kinetic friction refers to the resistive force that opposes the motion of a sliding object. It is calculated as the product of the coefficient of kinetic friction and the normal force acting on the object. In problems involving motion, balancing the applied forces with kinetic friction is essential to determine net motion or equilibrium conditions.

Magnetic Force on a Current-Carrying Conductor

This concept describes that a current-carrying conductor placed in a magnetic field experiences a force given by the equation F = I L × B. The cross product indicates that the force is perpendicular to both the current's direction and the magnetic field, and its magnitude is determined by the product of the current, the length of the conductor within the field, the magnetic field strength, and the sine of the angle between the current direction and the magnetic field.

Right-Hand Rule

The right-hand rule is a mnemonic used to determine the direction of the magnetic force acting on a current-carrying conductor. By pointing the thumb of the right hand in the direction of the current and the fingers in the direction of the magnetic field, the palm points in the direction of the magnetic force, providing a practical way to visualize the orientation of forces in electromagnetic problems.

Transcript

00:01 Hello friends here it is given.

00:02 A wire having linear mass density 1 gram per centimeter or 0 .1 kg per meter placed on a horizontal surface that having the coefficient of kinetic friction 0 .2 the wire carries the current of 1 .5 amp towards east.

00:57 What are the magnitude and direction of smallest magnetic field slides horizontally to the north moving towards north we have to find the magnetic field which enables the wire to move in this fashion force on the wire due to magnetic field we can calculate that will be ivl when the particle just move along the north direction then some of force along y -axis will be n plus fp cost of theta minus m g that must be zero see the figure if this is the direction of force just a moment this is the angle theta friction will be in this direction magnetic field is along this direction and this is the current through it and weight m g will act vertically downward so normal reaction upward so from here normal reaction we will get mg minus fp cost theta and friction will be mu into n that is mu k m g minus fv cost theta now submission of force along x -axis must be zero so f f sine of theta minus f must be 0 f v to be equal to mu m g minus f v cause of theta magnetic force we will get mu m g upon sine of theta plus mu cos theta for f f to be minimum d f v upon d theta must be zero so we can write d over d theta to be m g upon sine of theta plus mu cos theta simplifying for this you will write mu sine of theta to be equal to cost theta so theta will be 10 inverse of 1 upon mu and mu you know 0 .2 so it is to be 78 .7 degree so magnetic field will be fv upon il from equation 2 so this sorry equation 1 and 2 we can solve fv we have calculated mu m g upon sine of theta plus mu cos theta upon i into l m upon l can be written as mass per unit length mu lambda g upon sine of theta plus mu cos theta now substitute the value you will get mu is given 0 .2 g is 9 .8 .1 upon 1 .5 sorry i into l i forget to write i am just writing here length is just a moment i have to make a correction m upon n i have written mu so in denominator that there will be only i sign of 78 .7 plus 0 .2 cost of 78 .7...

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