A particle of charge 1.0 times 10^{-16} C mo | vedclass

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(D) The magnetic force on a charged particle moving in a magnetic field is given by the Lorentz force formula: $F = q(v 	imes B)$.Given:$q = 1.0 	im

Vedclass · Accepted Answer

$0.75$

Vedclass · Answer

(D) The magnetic force on a charged particle moving in a magnetic field is given by the Lorentz force formula: $F = q(v 	imes B)$.Given:$q = 1.0 	imes 10^{-16} \ C$$v = (2 \hat{i} + 4 \hat{j}) \ ms^{-1}$$B = B_0(\hat{i} + 4 \hat{j}) \ T$$F = 3 	imes 10^{-16} \hat{k} \ N$Substituting these values into the formula:$3 	imes 10^{-16} \hat{k} = 1.0 	imes 10^{-16} 	imes [(2 \hat{i} + 4 \hat{j}) 	imes B_0(\hat{i} + 4 \hat{j})]$$3 \hat{k} = B_0 	imes [2 \hat{i} 	imes \hat{i} + 8 \hat{i} 	imes \hat{j} + 4 \hat{j} 	imes \hat{i} + 16 \hat{j} 	imes \hat{j}]$Using cross product rules $(\hat{i} 	imes \hat{i} = 0, \hat{j} 	imes \hat{j} = 0, \hat{i} 	imes \hat{j} = \hat{k}, \hat{j} 	imes \hat{i} = -\hat{k})$:$3 \hat{k} = B_0 	imes [0 + 8 \hat{k} - 4 \hat{k} + 0]$$3 \hat{k} = B_0 	imes (4 \hat{k})$$4 B_0 = 3$$B_0 = \frac{3}{4} = 0.75 \ T$.

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