Question

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Harsh Pratap Singh :)

QUESTION :-
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Determine the resultant force acting on the wire frame shown placed in uniform magnetic field B directed into the the paper and perpendicular to its plane Given PQ=QR = l
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Note:-
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Answer should be easy to understand.

(You guys just Spammed my every question whatever questions I asked in last hours are all wrong except few.)
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Answer 1

Answer:

$\huge\underline\mathbb{HELLO}$

Let us consider a current carrying conductor of length L placed in a magnetic field B which is directed into the page .It is represented by X .

Let the field be confined to Length L.So only the part of wire of length L is inside the Magnetic field.

magnetic force on a single charge is given by:

F=qvB

let total charge inside the Magnetic field be Q so magnetic force on current carrying conductor is given by

F=QvB-----------1

the time taken by the charge q to cross the filed is

t=L/v

v=L/t-------------2

substituting this in equation 1 we get

F=Q(L/t)B

F=(Q/t) LB

but Q/T is electric current I

So F=ILB

This equation hold good when direction of electric current is perpendicular of magnetic field.

if at an angle then

F=ILBsinθ

We can use right hand rule to find out the direction of force on the current carrying wire. or Fleming's Left hand rule can also be used.

F will be maximum when I and B are perpendicular

F will be minimum when θ=0°

$\huge\underline\mathbb{HOPE IT HELPS}$

Answer 2

Answer:

Let us consider a current carrying conductor of length L placed in a magnetic field B which is directed into the page .It is represented by X .

Let the field be confined to Length L.So only the part of wire of length L is inside the Magnetic field.

magnetic force on a single charge is given by:

F=qvB

let total charge inside the Magnetic field be Q so magnetic force on current carrying conductor is given by

F=QvB-----------1

the time taken by the charge q to cross the filed is

t=L/v

v=L/t-------------2

substituting this in equation 1 we get

F=Q(L/t)B

F=(Q/t) LB

but Q/T is electric current I

So F=ILB

This equation hold good when direction of electric current is perpendicular of magnetic field.

if at an angle then

F=ILBsinθ

We can use right hand rule to find out the direction of force on the current carrying wire. or Fleming's Left hand rule can also be used.

F will be maximum when I and B are perpendicular

F will be minimum when θ=0°

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