JEE MAIN - Physics (2022 - 28th July Evening Shift - No. 12)
A triangular shaped wire carrying $$10 \mathrm{~A}$$ current is placed in a uniform magnetic field of $$0.5 \mathrm{~T}$$, as shown in figure. The magnetic force on segment $$\mathrm{CD}$$ is
(Given $$\mathrm{BC}=\mathrm{CD}=\mathrm{BD}=5 \mathrm{~cm}$$.)
_28th_July_Evening_Shift_en_12_1.png)
0.126 N
0.312 N
0.216 N
0.245 N
Explanation
$$\overrightarrow F = i\overrightarrow l \times \overrightarrow B $$
$$ = ilB\sin 60^\circ $$
$$ = 10 \times {5 \over {100}} \times 0.5 \times {{\sqrt 3 } \over 2}$$
$$ = 0.2165$$ N
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