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Question

# A wire lying along y - axis from y=0 to y=1 m carries a current of 2 mA in the negative y - direction. The wire lies in a non uniform magnetic field given by →B=(0.3 T/m)y^i+(0.4 T/m)y^j. The magnetic force on the entire wire will be:

A
3×104^j N
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B
6×103^k N
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C
3×104^k N
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D
3×104^k N
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Solution

## The correct option is D 3×10−4^k N Here By=0.4y and Bx=0.3y As shown in figure the length vector. →L=(−1 ^j) m Also, →Fm=i(→L×→B) gives the magnetic force experienced by wire. Due to y - component of magnetic field, the force will be zero, since →L×→B=0 in this case. →Fm=i[(→L)×(0.3y^i)] →F will be variable across the length of wire since y is changing. Let us assume a small current element of length dy, thus →dl=− dy ^j. Thus mangetic force on this segment will be, −→dF=i(→dl×→B) −→dF=i(−dy^j)×(0.3y^i) −→dF=−0.3iydy(−^k)=(0.3iydy) ^k Thus force on whole wire can be found by integrating and putting the limits y=0 to y=1 →F=∫y=1y=0d→F →F=0.3×(2×10−3)∫10ydy^k →F=0.6×10−3×[y22]10^k →F=0.3×10−3×(1−0)^k ∴→F=3×10−4 N ^k

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