First slide

Applications involving charged particles moving in a magnetic field

Question

The plane of a rectangular loop of wire with sides 0.05 m and 0.08 m is parallel to a uniform magnetic field of induction $1.5 \times 10^{- 2} T$ . A current of 10.0 ampere flow through the loop. If the side of length 0.08 m is normal and the side of length 0.05 m is parallel to the lines of induction, then the torque acting on the loop is

Moderate

A

6000 Nm
B

Zero
C

$1.2 \times 10^{- 2} N m$
D

$6 \times 10^{- 4} N m$

Solution

Torque $τ$ acting on a current carrying coil of area A placed in a magnetic field of
induction B is given by
$τ = N I B A \sin θ$
Where $I$ =current in the coil, $θ$ =angle which the normal to the plane of the coil makes with
the lines of induction B.
Here, $N = 1; B = 1.5 \times 10^{- 2} T$
$\begin{array}{l} A = 0.05 \times 0.08 = 40 \times 10^{- 4} m^{2} \\ I = 10.0 amp, θ = 90 ° = π / 2 \\ ∴ τ = (1.5 \times 10^{- 2}) (10.0) \times (1) (40 \times 10^{- 4}) \sin (π / 2) \\ = 6 \times 10^{- 4} Nm . \end{array}$

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