Question

A straight 2.20 m wire carries a typical household current of 1.50 A (in one direction) at a location where the earth's magnetic field is 0.550 gauss from south to north.

A.) Find the direction of the force that our planet's magnetic field exerts on this cord if is oriented so that the current in it is running from west to east. Please choose: a. Force is directed upward; b. Force is directed downward; c. There is no force exerting on the wire.; or d. The force is directed from east to west.

B.) Find the magnitude of the force that our planet's magnetic field exerts on this cord if is oriented so that the current in it is running from west to east.

C.) Find the direction of the force that our planet's magnetic field exerts on this cord if is oriented so that the current in it is running vertically upward. Please choose: a. The force is directed from east to west.; b. The force is directed from west to east.; c. The force is directed from north to south.; or d. The force is directed from south to north.

D.) Find the magnitude of the force that our planet's magnetic field exerts on this cord if is oriented so that the current in it is running vertically upward.

E.) Find the direction of the force that our planet's magnetic field exerts on this cord if is oriented so that the current in it is running from north to south. Please choose: a. Force is directed upward; b. Force is directed downward; c. There is no force exerting on the wire.; or d. The force is directed from north to south.

F.) Find the magnitude of the force that our planet's magnetic field exerts on this cord if is oriented so that the current in it is running from north to south.

G.) Is the magnetic force ever large enough to cause significant effects under normal household conditions? Yes or No.

Answer 1

Solution:-

A) We know that force exerted on a current carrying wire carrying a current (I) kept in a uniform magnetic field (B) of length(L) is given by( ) that is crossing the L(Vector) whose direction would be the direction of current  with B(Vector). As the current is running from west to east. So,the direction of L(vector) would be from west to east.

The magnetic vector(B) is from south to north. So, crossing the vector whose direction is from west to east with a vector whose direction is from south to north we will get the force vector in upward direction(a).

B) Magnitude of the force exerted on the current carrying wire would be = , where is the angle between L(vector) and B(vector).

Here, current is running from west to east so the direction of L(vector) is from west to east and The magnetic vector(B) is from south to north. So angle between the between L(vector) and B(vector) is 90⁰ and sin 90⁰=1.

So, the force exerted would be .

I = 1.50 A

B = 0.550 = 5.5 * 10^-5 Tesla

L = 2.2 meter

So, F=  1.50 *  2.2 * 5.5 * 10^-5 Newton

= 1.815 * 10^-4 Newton

C)   We know that force exerted on a current carrying wire carrying a current (I) kept in a uniform magnetic field (B) of length(L) is given by( ) that is crossing the L(Vector) whose direction would be the direction of current  with B(Vector). As the current is running vertically upward. So, So,the direction of L(vector) would be vertically upwards.

The magnetic vector(B) is from south to north. So, crossing the vector whose direction is vertically upward with a vector whose direction is from south to north we will get the force vector in east to west(a).

D)

Magnitude of the force exerted on the current carrying wire would be = , where is the angle between L(vector) and B(vector).

Here, current is running vertically upward so the direction of L(vector) is vertically upward and The magnetic vector(B) is from south to north. So angle between the between L(vector) and B(vector) is 90⁰ and sin 90⁰=1.

So, the force exerted would be >

I = 1.50 A

B = 0.550 = 5.5 * 10^-5 Tesla

L = 2.2 meter

So, F=  1.50 *  2.2 * 5.5 * 10^-5 Newton

= 1.815 * 10^-4 Newton