A current of 5 A is flowing in a wire of len | vedclass

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(A) The force $F$ on a current-carrying conductor in a uniform magnetic field is given by the formula: $F = B I L \sin 	heta$,where $B$ is the magnet

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$30$

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(A) The force $F$ on a current-carrying conductor in a uniform magnetic field is given by the formula: $F = B I L \sin 	heta$,where $B$ is the magnetic field,$I$ is the current,$L$ is the length of the wire,and $	heta$ is the angle between the magnetic field and the current direction.Given values are: $F = 7.5 \ N$,$B = 2 \ T$,$I = 5 \ A$,and $L = 1.5 \ m$.Substituting these values into the formula:$7.5 = 2 	imes 5 	imes 1.5 	imes \sin 	heta$$7.5 = 15 	imes \sin 	heta$$\sin 	heta = \frac{7.5}{15} = 0.5$Since $\sin 	heta = 0.5$,we have $	heta = 30^\circ$.

$A$ current of $5 \ A$ is flowing in a wire of length $1.5 \ m$. $A$ force of $7.5 \ N$ acts on it when it is placed in a uniform magnetic field of $2 \ T$. The angle between the magnetic field and the direction of the current is......$^o$

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